1
CHAPTER 1
Measurement
1-1 MEASURING THINGS, INCLUDING LENGTHS
Learning Objectives
After reading this module, you should be able to . . .
1.01 Identify the base quantities in the SI system.
1.02 Name the most frequently used prefixes for
SI units.
1.03 Change units (here for length, area, and volume) by
using chain-link conversions.
1.04 Explain that the meter is defined in terms of the speed of
light in vacuum.
Key Ideas
●Physics is based on measurement of physical quantities.
Certain physical quantities have been chosen as base quanti-
ties (such as length, time, and mass); each has been defined in
terms of a standard and given a unit of measure (such as meter,
second, and kilogram). Other physical quantities are defined in
terms of the base quantities and their standards and units.
●The unit system emphasized in this book is the International
System of Units (SI). The three physical quantities displayed
in Table 1-1 are used in the early chapters. Standards, which
must be both accessible and invariable, have been estab-
lished for these base quantities by international agreement.
These standards are used in all physical measurement, for
both the base quantities and the quantities derived from
them. Scientific notation and the prefixes of Table 1-2 are
used to simplify measurement notation.
●Conversion of units may be performed by using chain-link
conversions in which the original data are multiplied succes-
sively by conversion factors written as unity and the units are
manipulated like algebraic quantities until only the desired
units remain.
●The meter is defined as the distance traveled by light
during a precisely specified time interval.
What Is Physics?
Science and engineering are based on measurements and comparisons. Thus, we
need rules about how things are measured and compared, and we need
experiments to establish the units for those measurements and comparisons. One
purpose of physics (and engineering) is to design and conduct those experiments.
For example, physicists strive to develop clocks of extreme accuracy so that any
time or time interval can be precisely determined and compared. You may wonder
whether such accuracy is actually needed or worth the effort.Here is one example of
the worth: Without clocks of extreme accuracy, the Global Positioning System
(GPS) that is now vital to worldwide navigation would be useless.
Measuring Things
We discover physics by learning how to measure the quantities involved in
physics. Among these quantities are length, time, mass, temperature, pressure,
and electric current.
We measure each physical quantity in its own units, by comparison with a
standard. The unit is a unique name we assign to measures of that quantity—for
example, meter (m) for the quantity length. The standard corresponds to exactly
1.0 unit of the quantity. As you will see, the standard for length, which corresponds
to exactly 1.0 m, is the distance traveled by light in a vacuum during a certain
fraction of a second.We can define a unit and its standard in any way we care to.
However, the important thing is to do so in such a way that scientists around the
world will agree that our definitions are both sensible and practical.
Once we have set up a standard—say, for length—we must work out proce-
dures by which any length whatever, be it the radius of a hydrogen atom, the
wheelbase of a skateboard, or the distance to a star, can be expressed in terms of
the standard. Rulers, which approximate our length standard, give us one such
procedure for measuring length. However, many of our comparisons must be
indirect. You cannot use a ruler, for example, to measure the radius of an atom
or the distance to a star.
Base Quantities. There are so many physical quantities that it is a problem to
organize them. Fortunately, they are not all independent; for example, speed is the
ratio of a length to a time. Thus, what we do is pick out—by international agree-
ment—a small number of physical quantities, such as length and time, and assign
standards to them alone. We then define all other physical quantities in terms of
these base quantities and their standards (called base standards).Speed, for example,
is defined in terms of the base quantities length and time and their base standards.
Base standards must be both accessible and invariable. If we define the
length standard as the distance between one’s nose and the index finger on an
outstretched arm, we certainly have an accessible standard—but it will, of course,
vary from person to person.The demand for precision in science and engineering
pushes us to aim first for invariability. We then exert great effort to make dupli-
cates of the base standards that are accessible to those who need them.
The International System of Units
In 1971, the 14th General Conference on Weights and Measures picked seven
quantities as base quantities, thereby forming the basis of the International
System of Units, abbreviated SI from its French name and popularly known as
the metric system.Table 1-1 shows the units for the three base quantities—length,
mass, and time—that we use in the early chapters of this book. These units were
defined to be on a “human scale.”
Many SI derived units are defined in terms of these base units. For example,
the SI unit for power, called the watt (W), is defined in terms of the base units
for mass, length, and time.Thus, as you will see in Chapter 7,
1 watt 1W1 kg m2/s3, (1-1)
where the last collection of unit symbols is read as kilogram-meter squared per
second cubed.
To express the very large and very small quantities we often run into in
physics, we use scientific notation, which employs powers of 10. In this notation,
3 560 000 000 m 3.56 109m (1-2)
and 0.000 000 492 s 4.92 107s. (1-3)
Scientific notation on computers sometimes takes on an even briefer look, as in
3.56 E9 and 4.92 E–7, where E stands for “exponent of ten.” It is briefer still on
some calculators, where E is replaced with an empty space.
As a further convenience when dealing with very large or very small mea-
s
urements, we use the prefixes listed in Table 1-2. As you can see, each prefix
represents a certain power of 10, to be used as a multiplication factor. Attaching
a prefix to an SI unit has the effect of multiplying by the associated factor. Thus,
we can express a particular electric power as
1.27 109watts 1.27 gigawatts 1.27 GW (1-4)
2CHAPTER 1 MEASUREMENT
Table 1-1 Units for Three SI
Base Quantities
Quantity Unit Name Unit Symbol
Length meter m
Time second s
Mass kilogram kg
Table 1-2 Prefixes for SI Units
Factor PrefixaSymbol
1024 yotta- Y
1021 zetta- Z
1018 exa- E
1015 peta- P
1012 tera- T
109giga- G
106mega- M
103kilo- k
102hecto- h
101deka- da
101deci- d
102centi- c
103milli- m
106micro- m
109nano- n
1012 pico- p
1015 femto- f
1018 atto- a
1021 zepto- z
1024 yocto- y
aThe most frequently used prefixes are shown in
bold type.
or a particular time interval as
2.35 109s2.35 nanoseconds 2.35 ns. (1-5)
Some prefixes, as used in milliliter, centimeter, kilogram, and megabyte, are
probably familiar to you.
Changing Units
We often need to change the units in which a physical quantity is expressed. We
do so by a method called chain-link conversion. In this method, we multiply the
original measurement by a conversion factor (a ratio of units that is equal to
unity). For example, because 1 min and 60 s are identical time intervals, we have
Thus, the ratios (1 min)/(60 s) and (60 s)/(1 min) can be used as conversion
factors. This is not the same as writing or 60 1; each number and its unit
must be treated together.
Because multiplying any quantity by unity leaves the quantity unchanged, we
can introduce conversion factors wherever we find them useful. In chain-link
conversion, we use the factors to cancel unwanted units. For example, to convert
2 min to seconds, we have
(1-6)
If you introduce a conversion factor in such a way that unwanted units do not
cancel, invert the factor and try again. In conversions, the units obey the same
algebraic rules as variables and numbers.
Appendix D gives conversion factors between SI and other systems of units,
including non-SI units still used in the United States. However, the conversion
factors are written in the style of “1 min 60 s” rather than as a ratio. So, you
need to decide on the numerator and denominator in any needed ratio.
Length
In 1792, the newborn Republic of France established a new system of weights
and measures. Its cornerstone was the meter, defined to be one ten-millionth of
the distance from the north pole to the equator. Later, for practical reasons, this
Earth standard was abandoned and the meter came to be defined as the distance
between two fine lines engraved near the ends of a platinum–iridium bar, the
standard meter bar, which was kept at the International Bureau of Weights and
Measures near Paris. Accurate copies of the bar were sent to standardizing labo-
ratories throughout the world. These secondary standards were used to produce
other, still more accessible standards, so that ultimately every measuring device
derived its authority from the standard meter bar through a complicated chain
of comparisons.
Eventually, a standard more precise than the distance between two fine
scratches on a metal bar was required. In 1960, a new standard for the meter,
based on the wavelength of light, was adopted. Specifically, the standard for the
meter was redefined to be 1 650 763.73 wavelengths of a particular orange-red
light emitted by atoms of krypton-86 (a particular isotope, or type, of krypton) in
a gas discharge tube that can be set up anywhere in the world. This awkward
number of wavelengths was chosen so that the new standard would be close to
the old meter-bar standard.
2 min (2 min)(1) (2 min)
60 s
1 min
120 s.
1
60 1
1 min
60 s 1 and 60 s
1 min 1.
3
1-1 MEASURING THINGS, INCLUDING LENGTHS
By 1983, however, the demand for higher precision had reached such a point
that even the krypton-86 standard could not meet it, and in that year a bold step was
taken. The meter was redefined as the distance traveled by light in a specified time
interval.In the words of the 17th General Conference on Weights and Measures:
4CHAPTER 1 MEASUREMENT
The meter is the length of the path traveled by light in a vacuum during a time
interval of 1/299 792 458 of a second.
Table 1-3 Some Approximate Lengths
Measurement Length in Meters
Distance to the first
galaxies formed 2 1026
Distance to the
Andromeda galaxy 2 1022
Distance to the nearby
star Proxima Centauri 4 1016
Distance to Pluto 6 1012
Radius of Earth 6 106
Height of Mt. Everest 9 103
Thickness of this page 1 104
Length of a typical virus 1 108
Radius of a hydrogen atom 5 1011
Radius of a proton 1 1015
This time interval was chosen so that the speed of light cis exactly
c299 792 458 m/s.
Measurements of the speed of light had become extremely precise, so it made
sense to adopt the speed of light as a defined quantity and to use it to redefine
the meter.
Table 1-3 shows a wide range of lengths, from that of the universe (top line)
to those of some very small objects.
Significant Figures and Decimal Places
Suppose that you work out a problem in which each value consists of two digits.
Those digits are called significant figures and they set the number of digits that
you can use in reporting your final answer. With data given in two significant
figures, your final answer should have only two significant figures. However,
depending on the mode setting of your calculator, many more digits might be
displayed.Those extra digits are meaningless.
In this book, final results of calculations are often rounded to match the least
number of significant figures in the given data. (However, sometimes an extra
significant figure is kept.) When the leftmost of the digits to be discarded is 5 or
more, the last remaining digit is rounded up; otherwise it is retained as is. For
example, 11.3516 is rounded to three significant figures as 11.4 and 11.3279 is
rounded to three significant figures as 11.3. (The answers to sample problems in
this book are usually presented with the symbol instead of even if rounding
is involved.)
When a number such as 3.15 or 3.15 103is provided in a problem,the number
of significant figures is apparent, but how about the number 3000? Is it known to
only one significant figure (3 103)? Or is it known to as many as four significant
figures (3.000 103)? In this book, we assume that all the zeros in such given num-
bers as 3000 are significant, but you had better not make that assumption elsewhere.
Don’t confuse significant figures with decimal places. Consider the lengths
35.6 mm, 3.56 m, and 0.00356 m. They all have three significant figures but they
have one, two, and five decimal places, respectively.
ball’s builder most unhappy. Instead, because we want only
the nearest order of magnitude, we can estimate any quanti-
ties required in the calculation.
Calculations: Let us assume the ball is spherical with radius
R2 m. The string in the ball is not closely packed (there
are uncountable gaps between adjacent sections of string).
To allow for these gaps, let us somewhat overestimate
Sample Problem 1.01 Estimating order of magnitude, ball of string
The world’s largest ball of string is about 2 m in radius. To
the nearest order of magnitude, what is the total length L
of
the string in the ball?
KEY IDEA
We could, of course, take the ball apart and measure the to-
tal length L, but that would take great effort and make the
5
1-2 TIME
Additional examples, video, and practice available at WileyPLUS
1-2 TIME
Learning Objectives
After reading this module, you should be able to . . .
1.05 Change units for time by using chain-link conversions.
1.06 Use various measures of time, such as for motion or as
determined on different clocks.
Key Idea
●The second is defined in terms of the oscillations of light
emitted by an atomic (cesium-133) source. Accurate time
signals are sent worldwide by radio signals keyed to atomic
clocks in standardizing laboratories.
Time
Time has two aspects. For civil and some scientific purposes, we want to know
the time of day so that we can order events in sequence. In much scientific work,
we want to know how long an event lasts. Thus, any time standard must be able
to answer two questions: “When did it happen?” and “What is its duration?”
Table 1-4 shows some time intervals.
Any phenomenon that repeats itself is a possible time standard. Earth’s
rotation, which determines the length of the day, has been used in this way for
centuries; Fig. 1-1 shows one novel example of a watch based on that rotation.
A quartz clock, in which a quartz ring is made to vibrate continuously, can be
calibrated against Earth’s rotation via astronomical observations and used to
measure time intervals in the laboratory. However, the calibration cannot be
carried out with the accuracy called for by modern scientific and engineering
technology.
Table 1-4 Some Approximate Time Intervals
Time Interval
Measurement in Seconds
Lifetime of the
proton (predicted) 3 1040
Age of the universe 5 1017
Age of the pyramid of Cheops 1 1011
Human life expectancy 2 109
Length of a day 9 104
aThis is the earliest time after the big bang at which the laws of physics as we know them can be applied.
Time between human heartbeats 8 101
Lifetime of the muon 2 106
Shortest lab light pulse 1 1016
Lifetime of the most
unstable particle 1 1023
The Planck timea11043
Time Interval
Measurement in Seconds
the cross-sectional area of the string by assuming the
cross section is square, with an edge length d4 mm.
Then, with a cross-sectional area of d2and a length L, the
string occupies a total volume of
V(cross-sectional area)(length) d2L.
This is approximately equal to the volume of the ball, given
by , which is about 4R3because pis about 3. Thus, we
have the following
4
3
R3
d2L4R3,
or
2106m106m103km.
(Answer)
(Note that you do not need a calculator for such a simplified
calculation.) To the nearest order of magnitude, the ball
contains about 1000 km of string!
L4R3
d24(2 m)3
(4 103 m)2
Figure 1-1 When the metric system was
proposed in 1792, the hour was redefined
to provide a 10-hour day. The idea did not
catch on. The maker of this 10-hour watch
wisely provided a small dial that kept con-
ventional 12-hour time. Do the two dials
indicate the same time?
Steven Pitkin
Atomic clocks are so consistent that, in principle, two cesium clocks would have to
run for 6000 years before their readings would differ by more than 1 s. Even such
accuracy pales in comparison with that of clocks currently being developed; their
precision may be 1 part in 1018—that is, 1 s in 1 1018 s (which is about 3 1010 y).
6CHAPTER 1 MEASUREMENT
To meet the need for a better time standard, atomic clocks have
been developed. An atomic clock at the National Institute of
Standards and Technology (NIST) in Boulder, Colorado, is the stan-
dard for Coordinated Universal Time (UTC) in the United States. Its
time signals are available by shortwave radio (stations WWV and
WWVH) and by telephone (303-499-7111). Time signals (and related
information) are also available from the United States Naval
Observatory at website http://tycho.usno.navy.mil/time.html. (To set a
clock extremely accurately at your particular location, you would have
to account for the travel time required for these signals to reach you.)
Figure 1-2 shows variations in the length of one day on Earth over
a 4-year period, as determined by comparison with a cesium
(atomic) clock. Because the variation displayed by Fig. 1-2 is sea-
sonal and repetitious, we suspect the rotating Earth when there is a
difference between Earth and atom as timekeepers. The variation is
due to tidal effects caused by the Moon and to large-scale winds.
The 13th General Conference on Weights and Measures in 1967 adopted
a standard second based on the cesium clock:
One second is the time taken by 9 192 631 770 oscillations of the light (of a specified
wavelength) emitted by a cesium-133 atom.
Figure 1-2 Variations in the length of the
day over a 4-year period. Note that the
entire vertical scale amounts to only
3 ms (0.003 s).
1980 1981 1982 1983
+1
+2
+3
+4
Difference between length of
day and exactly 24 hours (ms)
1-3 MASS
Learning Objectives
After reading this module, you should be able to . . .
1.07 Change units for mass by using chain-link
conversions.
1.08 Relate density to mass and volume when the mass is
uniformly distributed.
Key Ideas
●The kilogram is defined in terms of a platinum–iridium
standard mass kept near Paris. For measurements on an
atomic scale, the atomic mass unit, defined in terms of
the atom carbon-12, is usually used.
●The density of a material is the mass per unit volume:
m
V.
Figure 1-3 The international 1 kg standard of
mass, a platinum–iridium cylinder 3.9 cm in
height and in diameter.
Mass
The Standard Kilogram
The SI standard of mass is a cylinder of
platinum and iridium (Fig. 1-3) that is kept
at the International Bureau of Weights
and Measures near Paris and assigned, by
Courtesy Bureau International des Poids et Me-
sures. Reproduced with permission of the BIPM.
7
1-3 MASS
international agreement, a mass of 1 kilogram. Accurate copies have been sent
to standardizing laboratories in other countries, and the masses of other bodies
can be determined by balancing them against a copy. Table 1-5 shows some
masses expressed in kilograms, ranging over about 83 orders of magnitude.
The U.S. copy of the standard kilogram is housed in a vault at NIST. It is
removed, no more than once a year, for the purpose of checking duplicate
copies that are used elsewhere. Since 1889, it has been taken to France twice for
recomparison with the primary standard.
A Second Mass Standard
The masses of atoms can be compared with one another more precisely than
they can be compared with the standard kilogram. For this reason, we have
a second mass standard. It is the carbon-12 atom, which, by international agree-
ment, has been assigned a mass of 12 atomic mass units (u).The relation between
the two units is
1u1.660 538 86 1027 kg, (1-7)
with an uncertainty of 10 in the last two decimal places. Scientists can, with
reasonable precision, experimentally determine the masses of other atoms rela-
tive to the mass of carbon-12. What we presently lack is a reliable means of
extending that precision to more common units of mass, such as a kilogram.
Density
As we shall discuss further in Chapter 14, density r(lowercase Greek letter rho)
is the mass per unit volume:
(1-8)
Densities are typically listed in kilograms per cubic meter or grams per cubic
centimeter.The density of water (1.00 gram per cubic centimeter) is often used as
a comparison. Fresh snow has about 10% of that density; platinum has a density
that is about 21 times that of water.
m
V.
Table 1-5 Some Approximate Masses
Mass in
Object Kilograms
Known universe 1 1053
Our galaxy 2 1041
Sun 2 1030
Moon 7 1022
Asteroid Eros 5 1015
Small mountain 1 1012
Ocean liner 7 107
Elephant 5 103
Grape 3 103
Speck of dust 7 1010
Penicillin molecule 5 1017
Uranium atom 4 1025
Proton 2 1027
Electron 9 1031
KEY IDEA
The density of the sand rsand in a sample is the mass per unit
volume—that is, the ratio of the total mass msand of the sand
grains to the total volume Vtotal of the sample:
(1-10)
Calculations: The total volume Vtotal of a sample is
Vtotal Vgrains Vvoids.
Substituting for Vvoids from Eq. 1-9 and solving for Vgrains
lead to
(1-11)Vgrains Vtotal
1e.
sand msand
Vtotal
.
Sample Problem 1.02 Density and liquefaction
A heavy object can sink into the ground during an earthquake
if the shaking causes the ground to undergo liquefaction, in
which the soil grains experience little friction as they slide
over one another. The ground is then effectively quicksand.
The possibility of liquefaction in sandy ground can be pre-
dicted in terms of the void ratio e for a sample of the ground:
(1-9)
Here, Vgrains is the total volume of the sand grains in the sam-
ple and Vvoids is the total volume between the grains (in the
voids). If eexceeds a critical value of 0.80, liquefaction can
occur during an earthquake.What is the corresponding sand
density rsand? Solid silicon dioxide (the primary component
of sand) has a density of 2.600 103kg/m3.
SiO2
eVvoids
Vgrains
.
Measurement in Physics Physics is based on measurement
of physical quantities. Certain physical quantities have been cho-
sen as base quantities (such as length, time, and mass); each has
been defined in terms of a standard and given a unit of measure
(such as meter, second, and kilogram). Other physical quantities
are defined in terms of the base quantities and their standards
and units.
SI Units The unit system emphasized in this book is the
International System of Units (SI). The three physical quantities
displayed in Table 1-1 are used in the early chapters. Standards,
which must be both accessible and invariable, have been estab-
lished for these base quantities by international agreement.
These standards are used in all physical measurement, for both
the base quantities and the quantities derived from them.
Scientific notation and the prefixes of Table 1-2 are used to sim-
plify measurement notation.
Changing Units Conversion of units may be performed by us-
ing chain-link conversions in which the original data are multiplied
successively by conversion factors written as unity and the units
are manipulated like algebraic quantities until only the desired
units remain.
Length The meter is defined as the distance traveled by light
during a precisely specified time interval.
Time The second is defined in terms of the oscillations of light
emitted by an atomic (cesium-133) source. Accurate time signals
are sent worldwide by radio signals keyed to atomic clocks in stan-
dardizing laboratories.
Mass The kilogram is defined in terms of a platinum–
iridium standard mass kept near Paris. For measurements on an
atomic scale, the atomic mass unit, defined in terms of the atom
carbon-12, is usually used.
Density The density r of a material is the mass per unit volume:
(1-8)
m
V.
Review & Summary
8CHAPTER 1 MEASUREMENT
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
Module 1-1 Measuring Things, Including Lengths
•1 Earth is approximately a sphere of radius 6.37 106m.
What are (a) its circumference in kilometers, (b) its surface area in
square kilometers, and (c) its volume in cubic kilometers?
•2 Agry is an old English measure for length, defined as 1/10 of a
line, where line is another old English measure for length, defined
as 1/12 inch. A common measure for length in the publishing busi-
ness is a point, defined as 1/72 inch. What is an area of 0.50 gry2in
points squared (points2)?
•3 The micrometer (1 mm) is often called the micron. (a) How
SSM
many microns make up 1.0 km? (b) What fraction of a centimeter
equals 1.0 mm? (c) How many microns are in 1.0 yd?
•4 Spacing in this book was generally done in units of points and
picas: 12 points 1 pica, and 6 picas 1 inch. If a figure was mis-
placed in the page proofs by 0.80 cm, what was the misplacement
in (a) picas and (b) points?
•5 Horses are to race over a certain English meadow
for a distance of 4.0 furlongs. What is the race distance in (a) rods
and (b) chains? (1 furlong 201.168 m, 1 rod 5.0292 m,
and 1 chain 20.117 m.)
WWWSSM
Additional examples, video, and practice available at WileyPLUS
From Eq. 1-8, the total mass msand of the sand grains is the
product of the density of silicon dioxide and the total vol-
ume of the sand grains:
(1-12)
Substituting this expression into Eq. 1-10 and then substitut-
ing for Vgrains from Eq. 1-11 lead to
(1-13)
sand
SiO2
V
total
V
total
1e
SiO2
1e.
msand
SiO2V
grains.
Substituting 2.600 103kg/m3and the critical value
of e0.80, we find that liquefaction occurs when the sand
density is less than
(Answer)
A building can sink several meters in such liquefaction.
sand 2.600 103 kg/m3
1.80 1.4 103 kg/m3.
SiO2
9
PROBLEMS
••6 You can easily convert common units and measures electroni-
cally, but you still should be able to use a conversion table, such as
those in Appendix D. Table 1-6 is part of a conversion table for a
system of volume measures once common in Spain; a volume of 1
fanega is equivalent to 55.501 dm3(cubic decimeters).To complete
the table, what numbers (to three significant figures) should be en-
tered in (a) the cahiz column, (b) the fanega column, (c) the cuar-
tilla column, and (d) the almude column, starting with the top
blank? Express 7.00 almudes in (e) medios, (f) cahizes, and (g) cu-
bic centimeters (cm3).
Table 1-6 Problem 6
cahiz fanega cuartilla almude medio
1 cahiz 1 12 48 144 288
1 fanega 1 4 12 24
1 cuartilla 136
1 almude 12
1 medio 1
••7 Hydraulic engineers in the United States often use, as a
unit of volume of water,the acre-foot, defined as the volume of wa-
ter that will cover 1 acre of land to a depth of 1 ft. A severe thun-
derstorm dumped 2.0 in. of rain in 30 min on a town of area 26
km2.What volume of water, in acre-feet, fell on the town?
••8 Harvard Bridge, which connects MIT with its fraternities
ILW
Module 1-2 Time
•10 Until 1883, every city and town in the United States kept its
own local time. Today, travelers reset their watches only when the
time change equals 1.0 h. How far, on the average, must you travel
in degrees of longitude between the time-zone boundaries at
which your watch must be reset by 1.0 h? (Hint: Earth rotates 360°
in about 24 h.)
•11 For about 10 years after the French Revolution, the French
government attempted to base measures of time on multiples of
ten: One week consisted of 10 days, one day consisted of 10 hours,
one hour consisted of 100 minutes, and one minute consisted of 100
seconds. What are the ratios of (a) the French decimal week to the
standard week and (b) the French decimal second to the standard
second?
•12 The fastest growing plant on record is a Hesperoyucca whip-
plei that grew 3.7 m in 14 days. What was its growth rate in micro-
meters per second?
•13 Three digital clocks A, B, and Crun at different rates and
3000 m
2000 km
Figure 1-5 Problem 9.
across the Charles River, has a length of 364.4 Smoots plus one
ear. The unit of one Smoot is based on the length of Oliver Reed
Smoot, Jr., class of 1962, who was carried or dragged length by
length across the bridge so that other pledge members of the
Lambda Chi Alpha fraternity could mark off (with paint)
1-Smoot lengths along the bridge.The marks have been repainted
biannually by fraternity pledges since the initial measurement,
usually during times of traffic congestion so that the police can-
not easily interfere. (Presumably, the police were originally up-
set because the Smoot is not an SI base unit, but these days they
seem to have accepted the unit.) Figure 1-4 shows three parallel
paths, measured in Smoots (S), Willies (W), and Zeldas (Z).
What is the length of 50.0 Smoots in (a) Willies and (b) Zeldas?
Figure 1-4 Problem 8.
••9 Antarctica is roughly semicircular, with a radius of 2000 km
(Fig. 1-5). The average thickness of its ice cover is 3000 m. How
many cubic centimeters of ice does Antarctica contain? (Ignore
the curvature of Earth.)
S
W
Z
032
60
212
258
216
0
do not have simultaneous readings of zero. Figure 1-6 shows si-
multaneous readings on pairs of the clocks for four occasions. (At
the earliest occasion, for example, Breads 25.0 s and Creads 92.0
s.) If two events are 600 s apart on clock A, how far apart are they
on (a) clock Band (b) clock C? (c) When clock Areads 400 s, what
does clock Bread? (d) When clock Creads 15.0 s,what does clock B
read? (Assume negative readings for prezero times.)
Figure 1-6 Problem 13.
•14 A lecture period (50 min) is close to 1 microcentury. (a) How
long is a microcentury in minutes? (b) Using
,
find the percentage difference from the approximation.
•15 A fortnight is a charming English measure of time equal to
2.0 weeks (the word is a contraction of “fourteen nights”).That is a
nice amount of time in pleasant company but perhaps a painful
string of microseconds in unpleasant company. How many mi-
croseconds are in a fortnight?
•16 Time standards are now based on atomic clocks. A promis-
ing second standard is based on pulsars, which are rotating neu-
tron stars (highly compact stars consisting only of neutrons).
Some rotate at a rate that is highly stable, sending out a radio
beacon that sweeps briefly across Earth once with each rotation,
like a lighthouse beacon. Pulsar PSR 1937 21 is an example; it
rotates once every 1.557 806 448 872 75 3 ms, where the trailing
3 indicates the uncertainty in the last decimal place (it does not
mean 3 ms). (a) How many rotations does PSR 1937 21 make
in 7.00 days? (b) How much time does the pulsar take to rotate ex-
actly one million times and (c) what is the associated uncertainty?
percentage difference
actual approximation
actual
100
A(s)
B(s)
C(s)
312 512
29020012525.0
92.0 142
10 CHAPTER 1 MEASUREMENT
•17 Five clocks are being tested in a laboratory. Exactly at
noon, as determined by the WWV time signal, on successive days
of a week the clocks read as in the following table. Rank the five
clocks according to their relative value as good timekeepers, best
to worst. Justify your choice.
Clock Sun. Mon. Tues. Wed. Thurs. Fri. Sat.
A12:36:40 12:36:56 12:37:12 12:37:27 12:37:44 12:37:59 12:38:14
B 11:59:59 12:00:02 11:59:57 12:00:07 12:00:02 11:59:56 12:00:03
C 15:50:45 15:51:43 15:52:41 15:53:39 15:54:37 15:55:35 15:56:33
D 12:03:59 12:02:52 12:01:45 12:00:38 11:59:31 11:58:24 11:57:17
E 12:03:59 12:02:49 12:01:54 12:01:52 12:01:32 12:01:22 12:01:12
••18 Because Earth’s rotation is gradually slowing, the length of
each day increases:The day at the end of 1.0 century is 1.0 ms longer
than the day at the start of the century. In 20 centuries, what is the
total of the daily increases in time?
•••19 Suppose that, while lying on a beach near the equator
watching the Sun set over a calm ocean, you start a stopwatch just
as the top of the Sun disappears. You then stand, elevating your
eyes by a height H1.70 m, and stop the watch when the top of
the Sun again disappears. If the elapsed time is t11.1 s, what is
the radius rof Earth?
Module 1-3 Mass
•20 The record for the largest glass bottle was set in 1992 by a
team in Millville, New Jersey—they blew a bottle with a volume of
193 U.S. fluid gallons. (a) How much short of 1.0 million cubic cen-
timeters is that? (b) If the bottle were filled with water at the
leisurely rate of 1.8 g/min, how long would the filling take? Water
has a density of 1000 kg/m3.
•21 Earth has a mass of 5.98 1024 kg.The average mass of the atoms
that make up Earth is 40 u. How many atoms are there in Earth?
•22 Gold, which has a density of 19.32 g/cm3, is the most ductile
metal and can be pressed into a thin leaf or drawn out into a long
fiber. (a) If a sample of gold, with a mass of 27.63 g, is pressed into
a leaf of 1.000 mm thickness, what is the area of the leaf? (b) If,
instead, the gold is drawn out into a cylindrical fiber of radius 2.500
mm, what is the length of the fiber?
•23 (a) Assuming that water has a density of exactly 1 g/cm3,
find the mass of one cubic meter of water in kilograms.
(b) Suppose that it takes 10.0 h to drain a container of 5700 m3of
water.What is the “mass flow rate,” in kilograms per second, of wa-
ter from the container?
••24 Grains of fine California beach sand are approximately
spheres with an average radius of 50 m and are made of silicon
dioxide, which has a density of 2600 kg/m3.What mass of sand grains
would have a total surface area (the total area of all the individual
spheres) equal to the surface area of a cube 1.00 m on an edge?
••25 During heavy rain, a section of a mountainside mea-
suring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m
deep slips into a valley in a mud slide.Assume that the mud ends up
uniformly distributed over a surface area of the valley measuring
0.40 km 0.40 km and that mud has a density of 1900 kg/m3. What
is the mass of the mud sitting above a 4.0 m2area of the valley floor?
••26 One cubic centimeter of a typical cumulus cloud contains
50 to 500 water drops, which have a typical radius of 10 mm. For
SSM
SSM that range, give the lower value and the higher value, respectively,
for the following. (a) How many cubic meters of water are in a
cylindrical cumulus cloud of height 3.0 km and radius 1.0 km? (b)
How many 1-liter pop bottles would that water fill? (c) Water has
a density of 1000 kg/m3. How much mass does the water in the
cloud have?
••27 Iron has a density of 7.87 g/cm3, and the mass of an iron atom
is 9.27 1026 kg. If the atoms are spherical and tightly packed, (a)
what is the volume of an iron atom and (b) what is the distance be-
tween the centers of adjacent atoms?
••28 A mole of atoms is 6.02 1023 atoms. To the nearest order
of magnitude, how many moles of atoms are in a large domestic
cat? The masses of a hydrogen atom, an oxygen atom, and a carbon
atom are 1.0 u, 16 u, and 12 u, respectively. (Hint: Cats are some-
times known to kill a mole.)
••29 On a spending spree in Malaysia, you buy an ox with
a weight of 28.9 piculs in the local unit of weights: 1 picul
100 gins, 1 gin 16 tahils, 1 tahil 10 chees, and 1 chee
10 hoons. The weight of 1 hoon corresponds to a mass of 0.3779 g.
When you arrange to ship the ox home to your astonished family,
how much mass in kilograms must you declare on the shipping
manifest? (Hint: Set up multiple chain-link conversions.)
••30 Water is poured into a container that has a small leak.
The mass mof the water is given as a function of time tby
m5.00t0.8 3.00t20.00, with t0, min grams, and tin sec-
onds. (a) At what time is the water mass greatest, and (b) what is
that greatest mass? In kilograms per minute, what is the rate of
mass change at (c) t2.00 s and (d) t5.00 s?
•••31 A vertical container with base area measuring 14.0 cm by
17.0 cm is being filled with identical pieces of candy, each with a
volume of 50.0 mm3and a mass of 0.0200 g.Assume that the volume
of the empty spaces between the candies is negligible. If the height
of the candies in the container increases at the rate of 0.250 cm/s, at
what rate (kilograms per minute) does the mass of the candies in
the container increase?
Additional Problems
32 In the United States, a doll house has the scale of 112 of a
real house (that is, each length of the doll house is that of the real
house) and a miniature house (a doll house to fit within a doll
house) has the scale of 1144 of a real house. Suppose a real house
(Fig. 1-7) has a front length of 20 m, a depth of 12 m, a height of 6.0 m,
and a standard sloped roof (vertical triangular faces on the ends)
of height 3.0 m. In cubic meters, what are the volumes of the corre-
sponding (a) doll house and (b) miniature house?
Figure 1-7 Problem 32.
6.0 m
12 m
20 m
3.0 m
1
12
11
PROBLEMS
33 A ton is a measure of volume frequently used in ship-
ping, but that use requires some care because there are at
least three types of tons: A displacement ton is equal to 7 barrels
bulk, a freight ton is equal to 8 barrels bulk, and a register ton is
equal to 20 barrels bulk. A barrel bulk is another measure of vol-
ume: 1 barrel bulk 0.1415 m3. Suppose you spot a shipping order
for “73 tons” of M&M candies, and you are certain that the client
who sent the order intended “ton” to refer to volume (instead of
weight or mass, as discussed in Chapter 5). If the client actually
meant displacement tons, how many extra U.S. bushels of the can-
dies will you erroneously ship if you interpret the order as (a) 73
freight tons and (b) 73 register tons? (1 m328.378 U.S.
bushels.)
34 Two types of barrel units were in use in the 1920s in the
United States.The apple barrel had a legally set volume of 7056 cu-
bic inches; the cranberry barrel, 5826 cubic inches. If a merchant
sells 20 cranberry barrels of goods to a customer who thinks he is
receiving apple barrels, what is the discrepancy in the shipment
volume in liters?
35 An old English children’s rhyme states, “Little Miss Muffet
sat on a tuffet, eating her curds and whey, when along came a spi-
der who sat down beside her. . . .”The spider sat down not because
of the curds and whey but because Miss Muffet had a stash of 11
tuffets of dried flies. The volume measure of a tuffet is given by
1 tuffet 2 pecks 0.50 Imperial bushel, where 1 Imperial bushel
36.3687 liters (L). What was Miss Muffet’s stash in (a) pecks,
(b) Imperial bushels, and (c) liters?
36 Table 1-7 shows some old measures of liquid volume. To
complete the table, what numbers (to three significant figures)
should be entered in (a) the wey column, (b) the chaldron column,
(c) the bag column, (d) the pottle column, and (e) the gill column,
starting from the top down? (f) The volume of 1 bag is equal to
0.1091 m3. If an old story has a witch cooking up some vile liquid
in a cauldron of volume 1.5 chaldrons, what is the volume in cubic
meters?
Table 1-7 Problem 36
wey chaldron bag pottle gill
1 wey 1 10/9 40/3 640 120 240
1 chaldron
1 bag
1 pottle
1 gill
37 A typical sugar cube has an edge length of 1 cm. If you had a
cubical box that contained a mole of sugar cubes, what would its
edge length be? (One mole 6.02 1023 units.)
38 An old manuscript reveals that a landowner in the time
of King Arthur held 3.00 acres of plowed land plus a live-
stock area of 25.0 perches by 4.00 perches. What was the total
area in (a) the old unit of roods and (b) the more modern unit of
square meters? Here, 1 acre is an area of 40 perches by 4 perches,
1 rood is an area of 40 perches by 1 perch, and 1 perch is the
length 16.5 ft.
39 A tourist purchases a car in England and ships it home to
the United States.The car sticker advertised that the car’s fuel con-
sumption was at the rate of 40 miles per gallon on the open road.
SSM
SSM The tourist does not realize that the U.K. gallon differs from the
U.S. gallon:
1 U.K. gallon 4.546 090 0 liters
1 U.S. gallon 3.785 411 8 liters.
For a trip of 750 miles (in the United States), how many gallons of
fuel does (a) the mistaken tourist believe she needs and (b) the car
actually require?
40 Using conversions and data in the chapter, determine
the number of hydrogen atoms required to obtain 1.0 kg of
hydrogen.A hydrogen atom has a mass of 1.0 u.
41 Acord is a volume of cut wood equal to a stack 8 ft
long, 4 ft wide, and 4 ft high. How many cords are in 1.0 m3?
42 One molecule of water (H2O) contains two atoms of hydrogen
and one atom of oxygen.A hydrogen atom has a mass of 1.0 u and an
atom of oxygen has a mass of 16 u, approximately. (a) What is the
mass in kilograms of one molecule of water? (b) How many mole-
cules of water are in the world’s oceans, which have an estimated total
mass of 1.4 1021 kg?
43 A person on a diet might lose 2.3 kg per week. Express the
mass loss rate in milligrams per second, as if the dieter could sense
the second-by-second loss.
44 What mass of water fell on the town in Problem 7? Water has
a density of 1.0 103kg/m3.
45 (a) A unit of time sometimes used in microscopic physics is
the shake. One shake equals 108s. Are there more shakes in a
second than there are seconds in a year? (b) Humans have ex-
isted for about 106years, whereas the universe is about 1010 years
old. If the age of the universe is defined as 1 “universe day,”
where a universe day consists of “universe seconds” as a normal
day consists of normal seconds, how many universe seconds have
humans existed?
46 A unit of area often used in measuring land areas is the hectare,
defined as 104m2. An open-pit coal mine consumes 75 hectares of
land, down to a depth of 26 m, each year. What volume of earth, in
cubic kilometers, is removed in this time?
47 An astronomical unit (AU) is the average distance
between Earth and the Sun, approximately 1.50 108km. The
speed of light is about 3.0 108m/s. Express the speed of light in
astronomical units per minute.
48 The common Eastern mole, a mammal, typically has a mass of
75 g, which corresponds to about 7.5 moles of atoms. (A mole of
atoms is 6.02 1023 atoms.) In atomic mass units (u), what is the
average mass of the atoms in the common Eastern mole?
49 A traditional unit of length in Japan is the ken (1 ken
1.97 m). What are the ratios of (a) square kens to square meters
and (b) cubic kens to cubic meters? What is the volume of a cylin-
drical water tank of height 5.50 kens and radius 3.00 kens in (c) cu-
bic kens and (d) cubic meters?
50 You receive orders to sail due east for 24.5 mi to put your sal-
vage ship directly over a sunken pirate ship. However, when your
divers probe the ocean floor at that location and find no evidence of
a ship, you radio back to your source of information, only to discover
that the sailing distance was supposed to be 24.5 nautical miles, not
regular miles. Use the Length table in Appendix D to calculate how
far horizontally you are from the pirate ship in kilometers.
SSM
SSM
12 CHAPTER 1 MEASUREMENT
51 The cubit is an ancient unit of length based on the distance
between the elbow and the tip of the middle finger of the mea-
surer. Assume that the distance ranged from 43 to 53 cm, and
suppose that ancient drawings indicate that a cylindrical pillar
was to have a length of 9 cubits and a diameter of 2 cubits. For
the stated range, what are the lower value and the upper value,
respectively, for (a) the cylinder’s length in meters, (b) the cylin-
der’s length in millimeters, and (c) the cylinder’s volume in cubic
meters?
52 As a contrast between the old and the modern and between
the large and the small, consider the following: In old rural
England 1 hide (between 100 and 120 acres) was the area of land
needed to sustain one family with a single plough for one year. (An
area of 1 acre is equal to 4047 m2.) Also, 1 wapentake was the area
of land needed by 100 such families. In quantum physics, the
cross-sectional area of a nucleus (defined in terms of the chance of
a particle hitting and being absorbed by it) is measured in units of
barns, where 1 barn is 1 1028 m2. (In nuclear physics jargon, if a
nucleus is “large,” then shooting a particle at it is like shooting a
bullet at a barn door, which can hardly be missed.) What is the
ratio of 25 wapentakes to 11 barns?
53 An astronomical unit (AU) is equal to the average
distance from Earth to the Sun, about 92.9 106mi. A parsec
(pc) is the distance at which a length of 1 AU would subtend an
angle of exactly 1 second of
arc (Fig. 1-8). A light-year (ly)
is the distance that light, trav-
eling through a vacuum with a
speed of 186 000 mi/s, would
cover in 1.0 year. Express the
Earth–Sun distance in (a)
parsecs and (b) light-years.
54 The description for a certain brand of house paint claims a cov-
erage of 460 ft2/gal. (a) Express this quantity in square meters per
liter. (b) Express this quantity in an SI unit (see Appendices A and
D). (c) What is the inverse of the original quantity, and (d) what is its
physical significance?
55 Strangely, the wine for a large wedding reception is to be
served in a stunning cut-glass receptacle with the interior dimen-
sions of 40 cm 40 cm 30 cm (height). The receptacle is to be
initially filled to the top. The wine can be purchased in bottles of
the sizes given in the following table. Purchasing a larger bottle in-
stead of multiple smaller bottles decreases the overall cost of the
wine. To minimize the cost, (a) which bottle sizes should be pur-
chased and how many of each should be purchased and, once the
receptacle is filled, how much wine is left over in terms of (b) stan-
dard bottles and (c) liters?
1 standard bottle
1 magnum 2 standard bottles
1 jeroboam 4 standard bottles
1 rehoboam 6 standard bottles
1 methuselah 8 standard bottles
1 salmanazar 12 standard bottles
1 balthazar 16 standard bottles 11.356 L
1 nebuchadnezzar 20 standard bottles
SSM
56 The corn–hog ratio is a financial term used in the pig market
and presumably is related to the cost of feeding a pig until it is
large enough for market. It is defined as the ratio of the market
price of a pig with a mass of 3.108 slugs to the market price of a
U.S. bushel of corn. (The word “slug” is derived from an old
German word that means “to hit”; we have the same meaning for
“slug” as a verb in modern English.) A U.S. bushel is equal to
35.238 L. If the corn–hog ratio is listed as 5.7 on the market ex-
change, what is it in the metric units of
(Hint:See the Mass table in Appendix D.)
57 You are to fix dinners for 400 people at a convention of
Mexican food fans. Your recipe calls for 2 jalapeño peppers per
serving (one serving per person). However, you have only ha-
banero peppers on hand. The spiciness of peppers is measured in
terms of the scoville heat unit (SHU). On average, one jalapeño
pepper has a spiciness of 4000 SHU and one habanero pepper has
a spiciness of 300 000 SHU. To get the desired spiciness, how many
habanero peppers should you substitute for the jalapeño peppers
in the recipe for the 400 dinners?
58 A standard interior staircase has steps each with a rise
(height) of 19 cm and a run (horizontal depth) of 23 cm. Research
suggests that the stairs would be safer for descent if the run were,
instead, 28 cm. For a particular staircase of total height 4.57 m, how
much farther into the room would the staircase extend if this
change in run were made?
59 In purchasing food for a political rally, you erroneously order
shucked medium-size Pacific oysters (which come 8 to 12 per U.S.
pint) instead of shucked medium-size Atlantic oysters (which
come 26 to 38 per U.S. pint). The filled oyster container shipped to
you has the interior measure of 1.0 m 12 cm 20 cm, and a U.S.
pint is equivalent to 0.4732 liter. By how many oysters is the order
short of your anticipated count?
60 An old English cookbook carries this recipe for cream of net-
tle soup: “Boil stock of the following amount: 1 breakfastcup plus
1 teacup plus 6 tablespoons plus 1 dessertspoon. Using gloves,
separate nettle tops until you have 0.5 quart; add the tops to the
boiling stock. Add 1 tablespoon of cooked rice and 1 saltspoon of
salt. Simmer for 15 min.” The following table gives some of the
conversions among old (premetric) British measures and among
common (still premetric) U.S. measures. (These measures just
scream for metrication.) For liquid measures, 1 British teaspoon
1 U.S. teaspoon. For dry measures, 1 British teaspoon 2 U.S. tea-
spoons and 1 British quart 1 U.S. quart. In U.S. measures, how
much (a) stock, (b) nettle tops, (c) rice, and (d) salt are required in
the recipe?
Old British Measures U.S. Measures
teaspoon 2 saltspoons tablespoon 3 teaspoons
dessertspoon 2 teaspoons half cup 8 tablespoons
tablespoon 2 dessertspoons cup 2 half cups
teacup 8 tablespoons
breakfastcup 2 teacups
price of 1 kilogram of pig
price of 1 liter of corn ?
An angle of
exactly 1 second
1 pc
1 AU
1
p
c
Figure 1-8 Problem 53.
CHAPTER 2
Motion Along a Straight Line
2-1 POSITION, DISPLACEMENT, AND AVERAGE VELOCITY
After reading this module, you should be able to …
2.01 Identify that if all parts of an object move in the same di-
rection and at the same rate, we can treat the object as if it
were a (point-like) particle. (This chapter is about the mo-
tion of such objects.)
2.02 Identify that the position of a particle is its location as
read on a scaled axis, such as an xaxis.
2.03 Apply the relationship between a particle’s
displacement and its initial and final positions.
2.04 Apply the relationship between a particle’s average
velocity, its displacement, and the time interval for that
displacement.
2.05 Apply the relationship between a particle’s average
speed, the total distance it moves, and the time interval for
the motion.
2.06 Given a graph of a particle’s position versus time,
determine the average velocity between any two particular
times.
●The position xof a particle on an xaxis locates the particle
with respect to the origin, or zero point, of the axis.
●The position is either positive or negative, according
to which side of the origin the particle is on, or zero if
the particle is at the origin. The positive direction on
an axis is the direction of increasing positive numbers;
the opposite direction is the negative direction on
the axis.
●The displacement xof a particle is the change in its
position:
●Displacement is a vector quantity. It is positive if the
particle has moved in the positive direction of the xaxis
and negative if the particle has moved in the negative
direction.
xx2x1.
●When a particle has moved from position x1to position x2
during a time interval tt2t1, its average velocity during
that interval is
.
●The algebraic sign of vavg indicates the direction of motion
(vavg is a vector quantity). Average velocity does not depend
on the actual distance a particle moves, but instead depends
on its original and final positions.
●On a graph of xversus t, the average velocity for a time in-
terval tis the slope of the straight line connecting the points
on the curve that represent the two ends of the interval.
●The average speed savg of a particle during a time interval t
depends on the total distance the particle moves in that time
interval:
savg total distance
t.
vavg x
tx2x1
t2t1
What Is Physics?
One purpose of physics is to study the motion of objects—how fast they move, for
example, and how far they move in a given amount of time. NASCAR engineers
are fanatical about this aspect of physics as they determine the performance of
their cars before and during a race. Geologists use this physics to measure
tectonic-plate motion as they attempt to predict earthquakes. Medical
researchers need this physics to map the blood flow through a patient when
diagnosing a partially closed artery, and motorists use it to determine how they
might slow sufficiently when their radar detector sounds a warning. There are
countless other examples. In this chapter, we study the basic physics of motion
where the object (race car, tectonic plate, blood cell, or any other object) moves
along a single axis. Such motion is called one-dimensional motion.
Key Ideas
Learning Objectives
13
Motion
The world, and everything in it, moves. Even seemingly stationary things, such as a
roadway, move with Earth’s rotation, Earth’s orbit around the Sun, the Sun’s orbit
around the center of the Milky Way galaxy, and that galaxy’s migration relative to
other galaxies.The classification and comparison of motions (called kinematics) is
often challenging.What exactly do you measure, and how do you compare?
Before we attempt an answer, we shall examine some general properties of
motion that is restricted in three ways.
1. The motion is along a straight line only.The line may be vertical, horizontal, or
slanted, but it must be straight.
2. Forces (pushes and pulls) cause motion but will not be discussed until
Chapter 5. In this chapter we discuss only the motion itself and changes in the
motion. Does the moving object speed up, slow down, stop, or reverse
direction? If the motion does change, how is time involved in the change?
3. The moving object is either a particle (by which we mean a point-like object
such as an electron) or an object that moves like a particle (such that every
portion moves in the same direction and at the same rate). A stiff pig slipping
down a straight playground slide might be considered to be moving like a par-
ticle; however,a tumbling tumbleweed would not.
Position and Displacement
To locate an object means to find its position relative to some reference point, of-
ten the origin (or zero point) of an axis such as the xaxis in Fig. 2-1.The positive
direction of the axis is in the direction of increasing numbers (coordinates), which
is to the right in Fig.2-1.The opposite is the negative direction.
For example, a particle might be located at x5 m, which means it is 5 m in
the positive direction from the origin. If it were at x5 m, it would be just as
far from the origin but in the opposite direction. On the axis, a coordinate of
5 m is less than a coordinate of 1 m, and both coordinates are less than a
coordinate of 5 m.A plus sign for a coordinate need not be shown, but a minus
sign must always be shown.
A change from position x1to position x2is called a displacement x, where
xx2x1. (2-1)
(The symbol , the Greek uppercase delta, represents a change in a quantity,
and it means the final value of that quantity minus the initial value.) When
numbers are inserted for the position values x1and x2in Eq. 2-1, a displacement
in the positive direction (to the right in Fig. 2-1) always comes out positive, and
a displacement in the opposite direction (left in the figure) always comes out
negative. For example, if the particle moves from x15m to x212 m, then
the displacement is x(12 m) (5 m) 7 m. The positive result indicates
that the motion is in the positive direction. If, instead, the particle moves from
x15m to x21 m, then x(1 m) (5 m) 4 m. The negative result in-
dicates that the motion is in the negative direction.
The actual number of meters covered for a trip is irrelevant; displacement in-
volves only the original and final positions. For example, if the particle moves
from x5 m out to x200 m and then back to x5 m, the displacement from
start to finish is x(5 m) (5 m) 0.
Signs. A plus sign for a displacement need not be shown, but a minus sign
must always be shown. If we ignore the sign (and thus the direction) of a displace-
ment, we are left with the magnitude (or absolute value) of the displacement. For
example, a displacement of x4 m has a magnitude of 4 m.
14 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Figure 2-1 Position is determined on an
axis that is marked in units of length (here
meters) and that extends indefinitely in
opposite directions. The axis name, here x,
is always on the positive side of the origin.
–3 0
Origin
–2 –1 1 2 3
Negative direction
Positive direction
x(m)
Displacement is an example of a vector quantity, which is a quantity that has
both a direction and a magnitude.We explore vectors more fully in Chapter 3, but
here all we need is the idea that displacement has two features: (1) Its magnitude
is the distance (such as the number of meters) between the original and final po-
sitions. (2) Its direction, from an original position to a final position, can be repre-
sented by a plus sign or a minus sign if the motion is along a single axis.
Here is the first of many checkpoints where you can check your understanding
with a bit of reasoning. The answers are in the back of the book.
15
2-1 POSITION, DISPLACEMENT, AND AVERAGE VELOCITY
Checkpoint 1
Here are three pairs of initial and final positions, respectively, along an xaxis.Which
pairs give a negative displacement: (a) 3m,5 m; (b) 3m,7 m; (c) 7 m, 3m?
Average Velocity and Average Speed
A compact way to describe position is with a graph of position xplotted as a func-
tion of time t—a graph of x(t). (The notation x(t) represents a function xof t, not
the product xtimes t.) As a simple example, Fig. 2-2 shows the position function
x(t) for a stationary armadillo (which we treat as a particle) over a 7 s time inter-
val.The animal’s position stays at x2m.
Figure 2-3 is more interesting, because it involves motion. The armadillo is
apparently first noticed at t0 when it is at the position x5 m. It moves
Figure 2-2 The graph of
x(t) for an armadillo that
is stationary at x2m.
The value of xis 2m
for all times t.
x(m)
t(s)
1234
+1
–1
–1
x(t)
0
This is a graph
of position x
versus time t
for a stationary
object.
Same position
for any time.
Figure 2-3 The graph of x(t) for a moving armadillo. The path associated with the graph is also shown, at three times.
x(m)
t(s)
1234
4
3
2
1
0
It is at position x = –5 m
when time t = 0 s.
Those data are plotted here.
This is a graph
of position x
versus time t
for a moving
object.
0
–5 2 x(m)
0 s
0
–5 2 x(m)
3 s
At x = 0 m when t = 3 s.
Plotted here.
At x = 2 m when t = 4 s.
Plotted here.
–1
–2
–3
–4
–5
x(t)0
–5 2 x(m)
4 s
A
16 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Figure 2-4 Calculation of the
average velocity between t1s
and t4 s as the slope of the line
that connects the points on the
x(t) curve representing those times.
The swirling icon indicates that a
figure is available in WileyPLUS
as an animation with voiceover.
x(m)
t(s)
x(t)
1234
4
3
2
1
–1
–2
–3
–4
–5
vavg = slope of this line
0
This horizontal distance is how long
it took, start to end:
Δt= 4 s – 1 s = 3 s
Start of interval
This vertical distance is how far
it moved, start to end:
Δx= 2 m – (–4 m) = 6 m
End of interval
Δx
__
Δt
rise
___
run
==
This is a graph
of position x
versus time t.
To find average velocity,
first draw a straight line,
start to end, and then
find the slope of the
line.
A
toward x0, passes through that point at t3 s, and then moves on to increas-
ingly larger positive values of x. Figure 2-3 also depicts the straight-line motion of
the armadillo (at three times) and is something like what you would see. The
graph in Fig.2-3 is more abstract, but it reveals how fast the armadillo moves.
Actually, several quantities are associated with the phrase “how fast.” One of
them is the average velocity vavg, which is the ratio of the displacement xthat
occurs during a particular time interval tto that interval:
(2-2)
The notation means that the position is x1at time t1and then x2at time t2.A com-
mon unit for vavg is the meter per second (m/s). You may see other units in the
problems, but they are always in the form of length/time.
Graphs. On a graph of xversus t,vavg is the slope of the straight line that
connects two particular points on the x(t) curve: one is the point that corresponds
to x2and t2, and the other is the point that corresponds to x1and t1. Like displace-
ment, vavg has both magnitude and direction (it is another vector quantity). Its
magnitude is the magnitude of the line’s slope. A positive vavg (and slope) tells us
that the line slants upward to the right; a negative vavg (and slope) tells us that the
line slants downward to the right. The average velocity vavg always has the same
sign as the displacement xbecause tin Eq. 2-2 is always positive.
Figure 2-4 shows how to find vavg in Fig. 2-3 for the time interval t1 s to t4s.
We draw the straight line that connects the point on the position curve at the begin-
ning of the interval and the point on the curve at the end of the interval.Then we find
the slope x/tof the straight line. For the given time interval, the average velocity is
Average speed savg is a different way of describing “how fast” a particle
moves. Whereas the average velocity involves the particle’s displacement x,the
average speed involves the total distance covered (for example, the number of
meters moved), independent of direction; that is,
(2-3)
Because average speed does not include direction, it lacks any algebraic sign.
Sometimes savg is the same (except for the absence of a sign) as vavg. However, the
two can be quite different.
savg total distance
t.
vavg 6 m
3 s 2 m/s.
vavg x
tx2x1
t2t1
.
17
2-1 POSITION, DISPLACEMENT, AND AVERAGE VELOCITY
Calculation: Here we find
(Answer)
To find vavg graphically, first we graph the function x(t) as
shown in Fig. 2-5, where the beginning and arrival points on
the graph are the origin and the point labeled as “Station.”Your
average velocity is the slope of the straight line connecting
those points; that is, vavg is the ratio of the rise (x10.4 km)
to the run (t0.62 h), which gives us vavg 16.8 km/h.
(d) Suppose that to pump the gasoline, pay for it, and walk
back to the truck takes you another 45 min. What is your
average speed from the beginning of your drive to your
return to the truck with the gasoline?
KEY IDEA
Your average speed is the ratio of the total distance you
move to the total time interval you take to make that move.
Calculation: The total distance is 8.4 km 2.0 km 2.0
km 12.4 km. The total time interval is 0.12 h 0.50 h
0.75 h 1.37 h.Thus, Eq. 2-3 gives us
(Answer)savg 12.4 km
1.37 h 9.1 km/h.
16.8 km/h 17 km/h.
vavg x
t10.4 km
0.62 h
Sample Problem 2.01 Average velocity, beat-up pickup truck
You drive a beat-up pickup truck along a straight road for
8.4 km at 70 km/h, at which point the truck runs out of gaso-
line and stops. Over the next 30 min, you walk another
2.0 km farther along the road to a gasoline station.
(a) What is your overall displacement from the beginning
of your drive to your arrival at the station?
KEY IDEA
Assume, for convenience, that you move in the positive di-
rection of an xaxis, from a first position of x10 to a second
position of x2at the station. That second position must be at
x28.4 km 2.0 km 10.4 km. Then your displacement x
along the xaxis is the second position minus the first position.
Calculation: From Eq. 2-1, we have
xx2x110.4 km 010.4 km. (Answer)
Thus, your overall displacement is 10.4 km in the positive
direction of the xaxis.
(b) What is the time interval tfrom the beginning of your
drive to your arrival at the station?
KEY IDEA
We already know the walking time interval twlk (0.50 h),
but we lack the driving time interval tdr. However, we
know that for the drive the displacement xdr is 8.4 km and
the average velocity vavg,dr is 70 km/h. Thus, this average
velocity is the ratio of the displacement for the drive to the
time interval for the drive.
Calculations: We first write
Rearranging and substituting data then give us
So,
(Answer)
(c) What is your average velocity vavg from the beginning of
your drive to your arrival at the station? Find it both numer-
ically and graphically.
KEY IDEA
From Eq. 2-2 we know that vavg for the entire trip is the ratio
of the displacement of 10.4 km for the entire trip to the time
interval of 0.62 h for the entire trip.
0.12 h 0.50 h 0.62 h.
ttdr twlk
tdr xdr
vavg,dr
8.4 km
70 km/h 0.12 h.
vavg,dr xdr
tdr
.
Additional examples, video, and practice available at WileyPLUS
Figure 2-5 The lines marked “Driving” and “Walking” are the
position–time plots for the driving and walking stages. (The plot
for the walking stage assumes a constant rate of walking.) The
slope of the straight line joining the origin and the point labeled
“Station” is the average velocity for the trip, from the beginning
to the station.
Position (km)
Time (h)
000.2 0.4 0.6
2
4
6
8
10
12
x
t
Walking
Driving
How far:
Δx = 10.4 km
Station
Driving ends, walking starts.
Slope of this
line gives
average
velocity.
How long:
Δt = 0.62 h
18 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Instantaneous Velocity and Speed
You have now seen two ways to describe how fast something moves: average
velocity and average speed, both of which are measured over a time interval t.
However, the phrase “how fast” more commonly refers to how fast a particle is
moving at a given instant—its instantaneous velocity (or simply velocity)v.
The velocity at any instant is obtained from the average velocity by shrinking
the time interval tcloser and closer to 0. As tdwindles, the average velocity
approaches a limiting value,which is the velocity at that instant:
(2-4)
Note that vis the rate at which position xis changing with time at a given instant;
that is, vis the derivative of xwith respect to t. Also note that vat any instant is
the slope of the position–time curve at the point representing that instant.
Velocity is another vector quantity and thus has an associated direction.
Speed is the magnitude of velocity; that is, speed is velocity that has been
stripped of any indication of direction, either in words or via an algebraic sign.
(Caution: Speed and average speed can be quite different.) A velocity of 5 m/s
and one of 5 m/s both have an associated speed of 5 m/s. The speedometer in a
car measures speed, not velocity (it cannot determine the direction).
vlim
t:0
x
tdx
dt .
2-2 INSTANTANEOUS VELOCITY AND SPEED
After reading this module, you should be able to . . .
2.07 Given a particle’s position as a function of time,
calculate the instantaneous velocity for any particular time.
2.08 Given a graph of a particle’s position versus time, deter-
mine the instantaneous velocity for any particular time.
2.09 Identify speed as the magnitude of the instantaneous
velocity.
●The instantaneous velocity (or simply velocity) vof a moving
particle is
where xx2x1and tt2t1.
vlim
t:0
x
tdx
dt ,
●The instantaneous velocity (at a particular time) may be
found as the slope (at that particular time) of the graph of x
versus t.
●Speed is the magnitude of instantaneous velocity.
Checkpoint 2
The following equations give the position x(t) of a particle in four situations (in each
equation, xis in meters, tis in seconds, and t0): (1) x3t2; (2) x4t22;
(3) x2/t2; and (4) x2. (a) In which situation is the velocity vof the particle con-
stant? (b) In which is vin the negative xdirection?
Calculations: The slope of x(t), and so also the velocity, is
zero in the intervals from 0 to 1 s and from 9 s on, so then
the cab is stationary. During the interval bc, the slope is con-
stant and nonzero, so then the cab moves with constant ve-
locity.We calculate the slope of x(t) then as
(2-5)
x
tv24 m 4.0 m
8.0 s 3.0 s 4.0 m/s.
Sample Problem 2.02 Velocity and slope of xversus t, elevator cab
Figure 2-6ais an x(t) plot for an elevator cab that is initially
stationary, then moves upward (which we take to be the pos-
itive direction of x), and then stops. Plot v(t).
KEY IDEA
We can find the velocity at any time from the slope of the
x(t) curve at that time.
Learning Objectives
Key Ideas
19
2-2 INSTANTANEOUS VELOCITY AND SPEED
Figure 2-6 (a) The x(t) curve for an elevator cab
that moves upward along an xaxis. (b)Thev(t)
curve for the cab. Note that it is the derivative
of the x(t) curve (vdx/dt). (c) The a(t) curve
for the cab. It is the derivative of the v(t) curve
(adv/dt). The stick figures along the bottom
suggest how a passenger’s body might feel dur-
ing the accelerations.
Additional examples, video, and practice available at WileyPLUS
The plus sign indicates that the cab is moving in the posi-
tive xdirection. These intervals (where v0 and v
4 m/s) are plotted in Fig. 2-6b. In addition, as the cab ini-
tially begins to move and then later slows to a stop,
vvaries as indicated in the intervals 1 s to 3 s and 8 s to 9 s.
Thus, Fig. 2-6bis the required plot. (Figure 2-6cis consid-
ered in Module 2-3.)
Given a v(t) graph such as Fig. 2-6b, we could “work
backward” to produce the shape of the associated x(t) graph
(Fig. 2-6a). However, we would not know the actual values
for xat various times, because the v(t) graph indicates
only changes in x. To find such a change in xduring any in-
Deceleration
Position (m)
Time (s)
t
0987654321
0
5
10
15
20
25
Slope
of x(t)
Δt
Δx
Velocity (m/s)
Time (s)
9
0
1
2
3
4
x
ab
cd
0
x(t)
v(t)
bc
da
v
t
0
–1
–2
–3
–4
1
2
t
Acceleration (m/s
2
)
(a)
(b)
(c)
987654321
87654321
a
a(t)
Acceleration
cb
ad
3
x = 24 m
at t= 8.0 s
x = 4.0 m
at t= 3.0 s
Slopes on the x versus t graph
are the values on the v versus t graph.
Slopes on the v versus t graph
are the values on the a versus t graph.
What you would feel.
terval, we must, in the language of calculus, calculate the
area “under the curve” on the v(t) graph for that interval.
For example, during the interval 3 s to 8 s in which the cab
has a velocity of 4.0 m/s, the change in xis
x(4.0 m/s)(8.0 s 3.0 s) 20 m. (2-6)
(This area is positive because the v(t) curve is above the
taxis.) Figure 2-6ashows that xdoes indeed increase by
20 m in that interval. However, Fig. 2-6bdoes not tell us the
values of xat the beginning and end of the interval. For that,
we need additional information, such as the value of xat
some instant.
Acceleration
When a particle’s velocity changes, the particle is said to undergo acceleration (or
to accelerate). For motion along an axis, the average acceleration aavg over a time
interval tis
(2-7)
where the particle has velocity v1at time t1and then velocity v2at time t2.The
instantaneous acceleration (or simply acceleration) is
(2-8)
In words, the acceleration of a particle at any instant is the rate at which its velocity
is changing at that instant. Graphically, the acceleration at any point is the slope of
the curve of v(t) at that point.We can combine Eq. 2-8 with Eq. 2-4 to write
(2-9)
In words, the acceleration of a particle at any instant is the second derivative of
its position x(t) with respect to time.
A common unit of acceleration is the meter per second per second: m/(s s)
or m/s2. Other units are in the form of length/(time time) or length/time2.
Acceleration has both magnitude and direction (it is yet another vector quan-
tity). Its algebraic sign represents its direction on an axis just as for displacement
and velocity; that is, acceleration with a positive value is in the positive direction
of an axis, and acceleration with a negative value is in the negative direction.
Figure 2-6 gives plots of the position, velocity, and acceleration of an ele-
vator moving up a shaft. Compare the a(t) curve with the v(t) curve—each
point on the a(t) curve shows the derivative (slope) of the v(t) curve at the
corresponding time. When vis constant (at either 0 or 4 m/s), the derivative is
zero and so also is the acceleration.When the cab first begins to move, the v(t)
adv
dt d
dt
dx
dt
d2x
dt 2.
adv
dt .
aavg v2v1
t2t1
v
t,
20 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
2-3 ACCELERATION
Learning Objectives
2.12 Given a graph of a particle’s velocity versus time, deter-
mine the instantaneous acceleration for any particular time
and the average acceleration between any two particular
times.
●Average acceleration is the ratio of a change in velocity v
to the time interval tin which the change occurs:
The algebraic sign indicates the direction of aavg.
aavg v
t.
●Instantaneous acceleration (or simply acceleration) ais the
first time derivative of velocity v(t)and the second time deriv-
ative of position x(t):
.
●On a graph of vversus t, the acceleration aat any time tis
the slope of the curve at the point that represents t.
adv
dt d2x
dt2
After reading this module, you should be able to . . .
2.10 Apply the relationship between a particle’s average
acceleration, its change in velocity, and the time interval
for that change.
2.11 Given a particle’s velocity as a function of time, calcu-
late the instantaneous acceleration for any particular time.
Key Ideas
curve has a positive derivative (the slope is positive), which means that a(t) is
positive. When the cab slows to a stop, the derivative and slope of the v(t)
curve are negative; that is, a(t) is negative.
Next compare the slopes of the v(t) curve during the two acceleration peri-
ods. The slope associated with the cab’s slowing down (commonly called “decel-
eration”) is steeper because the cab stops in half the time it took to get up to
speed. The steeper slope means that the magnitude of the deceleration is larger
than that of the acceleration, as indicated in Fig.2-6c.
Sensations. The sensations you would feel while riding in the cab of
Fig. 2-6 are indicated by the sketched figures at the bottom. When the cab first
accelerates, you feel as though you are pressed downward; when later the cab is
braked to a stop, you seem to be stretched upward. In between, you feel nothing
special. In other words, your body reacts to accelerations (it is an accelerometer)
but not to velocities (it is not a speedometer). When you are in a car traveling at
90 km/h or an airplane traveling at 900 km/h, you have no bodily awareness of
the motion. However, if the car or plane quickly changes velocity, you may be-
come keenly aware of the change, perhaps even frightened by it. Part of the thrill
of an amusement park ride is due to the quick changes of velocity that you un-
dergo (you pay for the accelerations, not for the speed).A more extreme example
is shown in the photographs of Fig. 2-7, which were taken while a rocket sled was
rapidly accelerated along a track and then rapidly braked to a stop.
g Units. Large accelerations are sometimes expressed in terms of gunits, with
1g9.8 m/s2(gunit). (2-10)
(As we shall discuss in Module 2-5, gis the magnitude of the acceleration of a
falling object near Earth’s surface.) On a roller coaster, you may experience brief
accelerations up to 3g, which is (3)(9.8 m/s2), or about 29 m/s2, more than enough
to justify the cost of the ride.
Signs. In common language, the sign of an acceleration has a nonscientific
meaning: positive acceleration means that the speed of an object is increasing, and
negative acceleration means that the speed is decreasing (the object is decelerat-
ing). In this book, however, the sign of an acceleration indicates a direction, not
21
2-3 ACCELERATION
Courtesy U.S. Air Force
Figure 2-7
Colonel J. P. Stapp in
a rocket sled as it is
brought up to high
speed (acceleration
out of the page) and
then very rapidly
braked (acceleration
into the page).
whether an object’s speed is increasing or decreasing. For example, if a car with an
initial velocity v25 m/s is braked to a stop in 5.0 s, then aavg 5.0 m/s2.The
acceleration is positive, but the car’s speed has decreased. The reason is the differ-
ence in signs: the direction of the acceleration is opposite that of the velocity.
Here then is the proper way to interpret the signs:
22 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
If the signs of the velocity and acceleration of a particle are the same, the speed
of the particle increases. If the signs are opposite, the speed decreases.
Checkpoint 3
A wombat moves along an xaxis.What is the sign of its acceleration if it is moving
(a) in the positive direction with increasing speed, (b) in the positive direction with
decreasing speed, (c) in the negative direction with increasing speed, and (d) in the
negative direction with decreasing speed?
Reasoning: We need to examine the expressions for x(t),
v(t), and a(t).
At t0, the particle is at x(0) 4 m and is moving
with a velocity of v(0) 27 m/s—that is, in the negative
direction of the xaxis. Its acceleration is a(0) 0 because just
then the particle’s velocity is not changing (Fig.2-8a).
For 0 t3 s, the particle still has a negative velocity,
so it continues to move in the negative direction. However,
its acceleration is no longer 0 but is increasing and positive.
Because the signs of the velocity and the acceleration are
opposite, the particle must be slowing (Fig.2-8b).
Indeed, we already know that it stops momentarily at
t3 s. Just then the particle is as far to the left of the origin
in Fig. 2-1 as it will ever get. Substituting t3 s into the
expression for x(t), we find that the particle’s position just
then is x50 m (Fig. 2-8c). Its acceleration is still positive.
For t3 s, the particle moves to the right on the axis.
Its acceleration remains positive and grows progressively
larger in magnitude. The velocity is now positive, and it too
grows progressively larger in magnitude (Fig.2-8d).
Sample Problem 2.03 Acceleration and dv/dt
A particle’s position on the xaxis of Fig. 2-1 is given by
x427tt3,
with xin meters and tin seconds.
(a) Because position xdepends on time t, the particle must
be moving. Find the particle’s velocity function v(t) and ac-
celeration function a(t).
KEY IDEAS
(1) To get the velocity function v(t), we differentiate the po-
sition function x(t) with respect to time. (2) To get the accel-
eration function a(t), we differentiate the velocity function
v(t) with respect to time.
Calculations: Differentiating the position function, we find
v27 3t2, (Answer)
with vin meters per second. Differentiating the velocity
function then gives us
a6t, (Answer)
with ain meters per second squared.
(b) Is there ever a time when v0?
Calculation: Setting v(t)0 yields
027 3t2,
which has the solution
t3 s. (Answer)
Thus, the velocity is zero both 3 s before and 3 s after the
clock reads 0.
(c) Describe the particle’s motion for t0. Figure 2-8 Four stages of the particle’s motion.
x
−50 m
t = 3 s
v = 0
a pos
reversing
(c)
t = 4 s
v pos
a pos
speeding up
(d)
0 4 m t = 0
v neg
a = 0
leftward
motion
(a)
t = 1 s
v neg
a pos
slowing
(b)
Additional examples, video, and practice available at WileyPLUS
Constant Acceleration: A Special Case
In many types of motion, the acceleration is either constant or approximately so.
For example, you might accelerate a car at an approximately constant rate when
a traffic light turns from red to green. Then graphs of your position, velocity,
and acceleration would resemble those in Fig. 2-9. (Note that a(t) in Fig. 2-9cis
constant, which requires that v(t) in Fig. 2-9bhave a constant slope.) Later when
you brake the car to a stop, the acceleration (or deceleration in common
language) might also be approximately constant.
Such cases are so common that a special set of equations has been derived
for dealing with them. One approach to the derivation of these equations is given
in this section. A second approach is given in the next section. Throughout both
sections and later when you work on the homework problems, keep in mind that
these equations are valid only for constant acceleration (or situations in which you
can approximate the acceleration as being constant).
First Basic Equation. When the acceleration is constant, the average accel-
eration and instantaneous acceleration are equal and we can write Eq. 2-7, with
some changes in notation, as
Here v0is the velocity at time t0 and vis the velocity at any later time t.We can
recast this equation as
vv0at. (2-11)
As a check, note that this equation reduces to vv0for t0, as it must.As a fur-
ther check, take the derivative of Eq. 2-11. Doing so yields dv/dt a, which is the
definition of a. Figure 2-9bshows a plot of Eq.2-11, the v(t) function; the function
is linear and thus the plot is a straight line.
Second Basic Equation. In a similar manner, we can rewrite Eq. 2-2 (with a
few changes in notation) as
vavg xx0
t0
aaavg vv0
t0.
23
2-4 CONSTANT ACCELERATION
2-4 CONSTANT ACCELERATION
After reading this module, you should be able to . . .
2.13 For constant acceleration, apply the relationships be-
tween position, displacement, velocity, acceleration, and
elapsed time (Table 2-1).
2.14 Calculate a particle’s change in velocity by integrating
its acceleration function with respect to time.
2.15 Calculate a particle’s change in position by integrating
its velocity function with respect to time.
●The following five equations describe the motion of a particle with constant acceleration:
These are not valid when the acceleration is not constant.
xx0vt 1
2at2.xx01
2(v0v)t,v2v0
22a(xx0),
xx0v0t1
2at2,vv0at,
Learning Objectives
Key Ideas
Figure 2-9 (a) The position x(t) of a particle
moving with constant acceleration. (b) Its
velocity v(t), given at each point by the
slope of the curve of x(t). (c) Its (constant)
acceleration, equal to the (constant) slope
of the curve of v(t).
0
a
Slope = 0
a(t)
(a)
(b)
(c)t
Position
0
x0
x
t
Slope varies
x(t)
Velocity
0
Slope = a
v(t)
v0
v
t
Acceleration
Slopes of the position graph
are plotted on the velocity graph.
Slope of the velocity graph is
plotted on the acceleration graph.
and then as
xx0vavgt, (2-12)
in which x0is the position of the particle at t0 and vavg is the average velocity
between t0 and a later time t.
For the linear velocity function in Eq. 2-11, the average velocity over any time
interval (say, from t0 to a later time t) is the average of the velocity at the be-
ginning of the interval (v0) and the velocity at the end of the interval (v). For
the interval from t0 to the later time tthen, the average velocity is
(2-13)
Substituting the right side of Eq. 2-11 for vyields, after a little rearrangement,
(2-14)
Finally, substituting Eq. 2-14 into Eq. 2-12 yields
(2-15)
As a check, note that putting t0 yields xx0, as it must. As a further check,
taking the derivative of Eq. 2-15 yields Eq. 2-11, again as it must. Figure 2-9a
shows a plot of Eq. 2-15; the function is quadratic and thus the plot is curved.
Three Other Equations. Equations 2-11 and 2-15 are the basic equations for
constant acceleration; they can be used to solve any constant acceleration prob-
lem in this book. However, we can derive other equations that might prove useful
in certain specific situations. First, note that as many as five quantities can possi-
bly be involved in any problem about constant acceleration—namely, xx0,v,t,
a, and v0. Usually, one of these quantities is not involved in the problem, either as
a given or as an unknown. We are then presented with three of the remaining
quantities and asked to find the fourth.
Equations 2-11 and 2-15 each contain four of these quantities, but not the
same four. In Eq. 2-11, the “missing ingredient” is the displacement xx0. In Eq.
2-15, it is the velocity v. These two equations can also be combined in three ways
to yield three additional equations, each of which involves a different “missing
variable.” First,we can eliminate tto obtain
(2-16)
This equation is useful if we do not know tand are not required to find it. Second,
we can eliminate the acceleration abetween Eqs. 2-11 and 2-15 to produce an
equation in which adoes not appear:
(2-17)
Finally, we can eliminate v0, obtaining
(2-18)
Note the subtle difference between this equation and Eq. 2-15. One involves the
initial velocity v0; the other involves the velocity vat time t.
Table 2-1 lists the basic constant acceleration equations (Eqs. 2-11 and 2-15) as
well as the specialized equations that we have derived.To solve a simple constant ac-
celeration problem, you can usually use an equation from this list (if you have the
list with you). Choose an equation for which the only unknown variable is the vari-
able requested in the problem.A simpler plan is to remember only Eqs. 2-11 and
2-15, and then solve them as simultaneous equations whenever needed.
xx0vt 1
2at 2.
xx01
2(v0v)t.
v2v0
22a(xx0).
xx0v0t1
2at 2.
vavg v01
2at.
vavg 1
2(v0v).
24 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Table 2-1 Equations for Motion with
Constant Accelerationa
Equation Missing
Number Equation Quantity
2-11 vv0at x x0
2-15 v
2-16 t
2-17 a
2-18 v0
aMake sure that the acceleration is indeed
constant before using the equations in this table.
xx0vt 1
2at2
xx01
2(v0v)t
v2v0
22a(xx0)
xx0v0t1
2at2
25
2-4 CONSTANT ACCELERATION
Checkpoint 4
The following equations give the position x(t) of a particle in four situations:(1) x
3t4; (2) x5t34t26; (3) x2/t24/t; (4) x5t23.To which of these
situations do the equations of Table 2-1 apply?
choose any initial numbers because we are looking for the
elapsed time, not a particular time in, say, the afternoon, but
let’s stick with these easy numbers.) We want the car to pass
the motorcycle, but what does that mean mathematically?
It means that at some time t, the side-by-side vehicles
are at the same coordinate: xcfor the car and the sum xm1
xm2for the motorcycle. We can write this statement mathe-
matically as
(2-19)
(Writing this first step is the hardest part of the problem.
That is true of most physics problems. How do you go from
the problem statement (in words) to a mathematical expres-
sion? One purpose of this book is for you to build up that
ability of writing the first step — it takes lots of practice just
as in learning,say, tae-kwon-do.)
Now let’s fill out both sides of Eq. 2-19, left side first. To
reach the passing point at xc, the car accelerates from rest. From
Eq. 2-15 , with x0and v0 0, we have
(2-20)
To write an expression for xm1for the motorcycle, we
first find the time tmit takes to reach its maximum speed vm,
using Eq. 2-11 (vv0 at). Substituting v0 0, v vm
58.8 m/s, and a am 8.40 m/s2, that time is
(2-21)
To get the distance xm1traveled by the motorcycle during
the first stage, we again use Eq. 2-15 with x0 0 and v0 0,
but we also substitute from Eq. 2-21 for the time.We find
(2-22)
For the remaining time of , the motorcycle travels
at its maximum speed with zero acceleration. To get the
distance, we use Eq. 2-15 for this second stage of the motion,
but now the initial velocity is (the speed at the endv0vm
ttm
xm11
2amtm
21
2am
vm
am
2
1
2
vm
2
am
.
58.8 m/s
8.40 m/s27.00 s.
tmvm
am
xc1
2act2.
(xx0v0t1
2at2)
xcxm1xm2.
Sample Problem 2.04 Drag race of car and motorcycle
A popular web video shows a jet airplane, a car, and a mo-
torcycle racing from rest along a runway (Fig. 2-10). Initially
the motorcycle takes the lead, but then the jet takes the lead,
and finally the car blows past the motorcycle. Here let’s focus
on the car and motorcycle and assign some reasonable values
to the motion. The motorcycle first takes the lead because its
(constant) acceleration am8.40 m/s2is greater than the car’s
(constant) acceleration ac5.60 m/s2, but it soon loses to the
car because it reaches its greatest speed vm58.8 m/s before
the car reaches its greatest speed vc106 m/s. How long does
the car take to reach the motorcycle?
KEY IDEAS
We can apply the equations of constant acceleration to both
vehicles, but for the motorcycle we must consider the mo-
tion in two stages: (1) First it travels through distance xm1
with zero initial velocity and acceleration am8.40 m/s2,
reaching speed vm58.8 m/s. (2) Then it travels through dis-
tance xm2with constant velocity vm58.8 m/s and zero ac-
celeration (that, too, is a constant acceleration). (Note that
we symbolized the distances even though we do not know
their values. Symbolizing unknown quantities is often help-
ful in solving physics problems, but introducing such un-
knowns sometimes takes physics courage.)
Calculations: So that we can draw figures and do calcula-
tions, let’s assume that the vehicles race along the positive di-
rection of an xaxis, starting from x0 at time t0. (We can
Figure 2-10 A jet airplane, a car, and a motorcycle just after
accelerating from rest.
of the first stage) and the acceleration is a 0. So, the dis-
tance traveled during the second stage is
(2-23)xm2vm(ttm)vm(t7.00 s).
Additional examples, video, and practice available at WileyPLUS
Another Look at Constant Acceleration*
The first two equations in Table 2-1 are the basic equations from which the others
are derived. Those two can be obtained by integration of the acceleration with
the condition that ais constant. To find Eq. 2-11, we rewrite the definition of ac-
celeration (Eq. 2-8) as
dv a dt.
We next write the indefinite integral (or antiderivative) of both sides:
Since acceleration ais a constant, it can be taken outside the integration.We obtain
or vat C. (2-25)
To evaluate the constant of integration C, we let t0, at which time vv0.
Substituting these values into Eq. 2-25 (which must hold for all values of t,
including t0) yields
v0(a)(0) CC.
Substituting this into Eq. 2-25 gives us Eq. 2-11.
To derive Eq. 2-15, we rewrite the definition of velocity (Eq. 2-4) as
dx v dt
and then take the indefinite integral of both sides to obtain
dx vdt.
dv adt
dv adt.
26 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
that at t7.00 s the plot for the motorcycle switches from
being curved (because the speed had been increasing) to be-
ing straight (because the speed is thereafter constant).
Figure 2-11 Graph of position versus time for car and motorcycle.
*This section is intended for students who have had integral calculus.
1000
800
600
400
200
0
Acceleration
ends
Motorcycle
Car
Car passes
motorcycle
x (m)
t (s)
0 5 10 15 20
To finish the calculation, we substitute Eqs. 2-20, 2-22, and
2-23 into Eq. 2-19, obtaining
(2-24)
This is a quadratic equation. Substituting in the given data,
we solve the equation (by using the usual quadratic-equa-
tion formula or a polynomial solver on a calculator), finding
t4.44 s and t16.6 s.
But what do we do with two answers? Does the car pass
the motorcycle twice? No, of course not, as we can see in the
video. So, one of the answers is mathematically correct but
not physically meaningful. Because we know that the car
passes the motorcycle after the motorcycle reaches its maxi-
mum speed at t7.00 s, we discard the solution with t
7.00 s as being the unphysical answer and conclude that the
passing occurs at
(Answer)
Figure 2-11 is a graph of the position versus time for
the two vehicles, with the passing point marked. Notice
t16.6 s.
1
2act21
2
vm
2
am
vm(t7.00 s).
27
2-5 FREE-FALL ACCELERATION
Next, we substitute for vwith Eq. 2-11:
Since v0is a constant, as is the acceleration a, this can be rewritten as
Integration now yields
(2-26)
where Cis another constant of integration. At time t0, we have xx0.
Substituting these values in Eq. 2-26 yields x0C. Replacing Cwith x0in Eq.
2-26 gives us Eq. 2-15.
xv0t1
2at 2C,
dx v0dt atdt.
dx (v0at)dt.
2-5 FREE-FALL ACCELERATION
After reading this module, you should be able to . . .
2.16 Identify that if a particle is in free flight (whether
upward or downward) and if we can neglect the
effects of air on its motion, the particle has a constant
downward acceleration with a magnitude gthat we take to
be 9.8 m/s2.
2.17 Apply the constant-acceleration equations (Table 2-1) to
free-fall motion.
●An important example of straight-line motion with constant
acceleration is that of an object rising or falling freely near
Earth’s surface. The constant acceleration equations de-
scribe this motion, but we make two changes in notation:
(1) we refer the motion to the vertical yaxis with yvertically
up; (2) we replace awith g, where gis the magnitude of the
free-fall acceleration. Near Earth’s surface,
g9.8 m/s232 ft/s2.
Learning Objectives
Key Ideas
Free-Fall Acceleration
If you tossed an object either up or down and could somehow eliminate the
effects of air on its flight, you would find that the object accelerates downward at
a certain constant rate.That rate is called the free-fall acceleration, and its magni-
tude is represented by g. The acceleration is independent of the object’s charac-
teristics, such as mass, density, or shape; it is the same for all objects.
Two examples of free-fall acceleration are shown in Fig. 2-12, which is a series
of stroboscopic photos of a feather and an apple. As these objects fall, they
accelerate downward—both at the same rate g.Thus, their speeds increase at the
same rate, and they fall together.
The value of gvaries slightly with latitude and with elevation. At sea level
in Earth’s midlatitudes the value is 9.8 m/s2(or 32 ft/s2), which is what you
should use as an exact number for the problems in this book unless otherwise
noted.
The equations of motion in Table 2-1 for constant acceleration also apply to
free fall near Earth’s surface; that is, they apply to an object in vertical flight,
either up or down, when the effects of the air can be neglected. However, note
that for free fall: (1) The directions of motion are now along a vertical yaxis
instead of the xaxis, with the positive direction of yupward. (This is important
for later chapters when combined horizontal and vertical motions are examined.)
(2) The free-fall acceleration is negative—that is, downward on the yaxis, toward
Earth’s center—and so it has the value gin the equations.
Figure 2-12 A feather and an apple free
fall in vacuum at the same magnitude of
acceleration g. The acceleration increases
the distance between successive images. In
the absence of air, the feather and apple
fall together.
© Jim Sugar/CORBIS
28 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Checkpoint 5
(a) If you toss a ball straight up,what is the sign of the ball’s displacement for the ascent,
from the release point to the highest point? (b) What is it for the descent, from the high-
est point back to the release point? (c) What is the ball’s acceleration at its highest point?
or
If we temporarily omit the units (having noted that they are
consistent), we can rewrite this as
4.9t212t5.0 0.
Solving this quadratic equation for tyields
t0.53 s and t1.9 s. (Answer)
There are two such times! This is not really surprising
because the ball passes twice through y5.0 m, once on the
way up and once on the way down.
5.0 m (12 m/s)t(1
2)(9.8 m/s2)t2.
Sample Problem 2.05 Time for full up-down flight, baseball toss
In Fig. 2-13, a pitcher tosses a baseball up along a yaxis, with
an initial speed of 12 m/s.
(a) How long does the ball take to reach its maximum height?
KEY IDEAS
(1) Once the ball leaves the pitcher and before it returns to
his hand, its acceleration is the free-fall acceleration ag.
Because this is constant, Table 2-1 applies to the motion.
(2) The velocity vat the maximum height must be 0.
Calculation: Knowing v,a, and the initial velocity
v012 m/s, and seeking t, we solve Eq. 2-11, which contains
those four variables.This yields
(Answer)
(b) What is the ball’s maximum height above its release point?
Calculation: We can take the ball’s release point to be y00.
We can then write Eq.2-16 in ynotation, set yy0yand v
0 (at the maximum height),and solve for y.We get
(Answer)
(c) How long does the ball take to reach a point 5.0 m above
its release point?
Calculations: We know v0,ag, and displacement y
y05.0 m, and we want t, so we choose Eq. 2-15. Rewriting
it for yand setting y00 give us
yv0t1
2gt2,
yv2v0
2
2a0(12 m/s)2
2(9.8 m/s2)7.3 m.
tvv0
a012 m/s
9.8 m/s21.2 s.
Figure 2-13 A pitcher tosses a
baseball straight up into the air.
The equations of free fall apply
for rising as well as for falling
objects, provided any effects
from the air can be neglected.
Ball
y = 0
y
v = 0 at
highest point
During ascent,
a = –g,
speed decreases,
and velocity
becomes less
positive
During
descent,
a = –g,
speed
increases,
and velocit
y
becomes
more
negative
Suppose you toss a tomato directly upward with an initial (positive) velocity v0
and then catch it when it returns to the release level. During its free-fall flight (from
just after its release to just before it is caught), the equations of Table 2-1 apply to its
motion. The acceleration is always ag9.8 m/s2, negative and thus down-
ward. The velocity, however, changes, as indicated by Eqs. 2-11 and 2-16: during the
ascent, the magnitude of the positive velocity decreases, until it momentarily be-
comes zero. Because the tomato has then stopped, it is at its maximum height.
During the descent,the magnitude of the (now negative) velocity increases.
Additional examples, video, and practice available at WileyPLUS
The free-fall acceleration near Earth’s surface is ag9.8 m/s2, and the
magnitude of the acceleration is g9.8 m/s2. Do not substitute 9.8 m/s2for g.
29
2-6 GRAPHICAL INTEGRATION IN MOTION ANALYSIS
2-6 GRAPHICAL INTEGRATION IN MOTION ANALYSIS
After reading this module, you should be able to . . .
2.18 Determine a particle’s change in velocity by graphical
integration on a graph of acceleration versus time.
2.19 Determine a particle’s change in position by graphical
integration on a graph of velocity versus time.
●On a graph of acceleration aversus time t, the change in
the velocity is given by
The integral amounts to finding an area on the graph:
t1
t0
adt
area between acceleration curve
and time axis, from t0to t1
.
v1v0t1
t0
adt.
●On a graph of velocity vversus time t, the change in the
position is given by
where the integral can be taken from the graph as
t1
t0
vdt
area between velocity curve
and time axis, from t0to t1
.
x1x0t1
t0
vdt,
Learning Objectives
Key Ideas
Graphical Integration in Motion Analysis
Integrating Acceleration. When we have a graph of an object’s acceleration aver-
sus time t, we can integrate on the graph to find the velocity at any given time.
Because ais defined as adv/dt,the Fundamental Theorem of Calculus tells us that
(2-27)
The right side of the equation is a definite integral (it gives a numerical result rather
than a function),v0is the velocity at time t0,and v1is the velocity at later time t1.The def-
inite integral can be evaluated from an a(t) graph, such as in Fig.2-14a.In particular,
(2-28)
If a unit of acceleration is 1 m/s2and a unit of time is 1 s, then the correspon-
ding unit of area on the graph is
(1 m/s2)(1 s) 1 m/s,
which is (properly) a unit of velocity.When the acceleration curve is above the time
axis, the area is positive; when the curve is below the time axis, the area is negative.
Integrating Velocity. Similarly, because velocity vis defined in terms of the posi-
tion xas vdx/dt,then
(2-29)
where x0is the position at time t0and x1is the position at time t1. The definite
integral on the right side of Eq. 2-29 can be evaluated from a v(t) graph, like that
shown in Fig.2-14b. In particular,
(2-30)
If the unit of velocity is 1 m/s and the unit of time is 1 s, then the corre-
sponding unit of area on the graph is
(1 m/s)(1 s) 1m,
which is (properly) a unit of position and displacement.Whether this area is posi-
tive or negative is determined as described for the a(t) curve of Fig.2-14a.
t1
t0
vdt
area between velocity curve
and time axis, from t0to t1
.
x1x0t1
t0
vdt,
t1
t0
adt
area between acceleration curve
and time axis, from t0to t1
.
v1v0t1
t0
adt.
Figure 2-14 The area between a plotted
curve and the horizontal time axis, from
time t0to time t1, is indicated for (a) a
graph of acceleration aversus tand (b) a
graph of velocity vversus t.
a
t0t
t1
Area
(a)
v
t0t
t1
Area
(b)
This area gives the
change in velocity.
This area gives the
change in position.
30 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Combining Eqs. 2-27 and 2-28, we can write
(2-31)
For convenience, let us separate the area into three regions
(Fig. 2-15b).From 0 to 40 ms, region Ahas no area:
areaA0.
From 40 ms to 100 ms,region Bhas the shape of a triangle,with
area
From 100 ms to 110 ms, region Chas the shape of a rectan-
gle, with area
areaC(0.010 s)(50 m/s2)0.50 m/s.
Substituting these values and v00 into Eq.2-31 gives us
v1001.5 m/s 0.50 m/s,
or v12.0 m/s 7.2 km/h. (Answer)
Comments: When the head is just starting to move forward,
the torso already has a speed of 7.2 km/h. Researchers argue
that it is this difference in speeds during the early stage of a
rear-end collision that injures the neck. The backward whip-
ping of the head happens later and could, especially if there is
no head restraint,increase the injury.
areaB1
2(0.060 s)(50 m/s2)1.5 m/s.
v1v0
area between acceleration curve
and time axis, from t0 to t1
.
Sample Problem 2.06 Graphical integration aversus t, whiplash injury
“Whiplash injury” commonly occurs in a rear-end collision
where a front car is hit from behind by a second car. In the
1970s, researchers concluded that the injury was due to the
occupant’s head being whipped back over the top of the seat
as the car was slammed forward. As a result of this finding,
head restraints were built into cars, yet neck injuries in rear-
end collisions continued to occur.
In a recent test to study neck injury in rear-end collisions,
a volunteer was strapped to a seat that was then moved
abruptly to simulate a collision by a rear car moving at
10.5 km/h. Figure 2-15agives the accelerations of the volun-
teer’s torso and head during the collision, which began at time
t0. The torso acceleration was delayed by 40 ms because
during that time interval the seat back had to compress
against the volunteer. The head acceleration was delayed by
an additional 70 ms. What was the torso speed when the head
began to accelerate?
KEY IDEA
We can calculate the torso speed at any time by finding an
area on the torso a(t) graph.
Calculations: We know that the initial torso speed is v00
at time t00, at the start of the “collision.” We want the
torso speed v1at time t1110 ms, which is when the head
begins to accelerate.
Figure 2-15 (a) The a(t) curve of the torso and head of a volunteer
in a simulation of a rear-end collision. (b) Breaking up the region
between the plotted curve and the time axis to calculate the area.
Additional examples, video, and practice available at WileyPLUS
Position The position x of a particle on an xaxis locates the par-
ticle with respect to the origin, or zero point, of the axis.The position
is either positive or negative, according to which side of the origin
the particle is on, or zero if the particle is at the origin. The positive
direction on an axis is the direction of increasing positive numbers;
the opposite direction is the negative direction on the axis.
Displacement The displacement xof a particle is the change
in its position:
xx2x1. (2-1)
Displacement is a vector quantity. It is positive if the particle has
moved in the positive direction of the xaxis and negative if the
particle has moved in the negative direction.
Review & Summary
Average Velocity When a particle has moved from position x1
to position x2during a time interval tt2t1, its average velocity
during that interval is
(2-2)
The algebraic sign of vavg indicates the direction of motion (vavg is a
vector quantity). Average velocity does not depend on the actual
distance a particle moves, but instead depends on its original and
final positions.
On a graph of xversus t, the average velocity for a time interval
tis the slope of the straight line connecting the points on the curve
that represent the two ends of the interval.
vavg x
tx2x1
t2t1
.
a (m/s2)
Head
0 40 80 120 160
50
100
t (ms)
(a)
Torso
50
(b)A
BC
a
t
40 100 110
The total area gives the
change in velocity.
31
QUESTIONS
Average Speed The average speed savg of a particle during a
time interval tdepends on the total distance the particle moves in
that time interval:
(2-3)
Instantaneous Velocity The instantaneous velocity (or sim-
ply velocity)vof a moving particle is
(2-4)
where xand tare defined by Eq. 2-2.The instantaneous velocity
(at a particular time) may be found as the slope (at that particular
time) of the graph of xversus t.Speed is the magnitude of instanta-
neous velocity.
Average Acceleration Average acceleration is the ratio of a
change in velocity vto the time interval tin which the change occurs:
(2-7)
The algebraic sign indicates the direction of aavg.
Instantaneous Acceleration Instantaneous acceleration (or
simply acceleration)ais the first time derivative of velocity v(t)
aavg v
t.
vlim
t:0
x
tdx
dt ,
savg total distance
t.
and the second time derivative of position x(t):
(2-8, 2-9)
On a graph of vversus t, the acceleration aat any time tis the slope
of the curve at the point that represents t.
Constant Acceleration The five equations in Table 2-1
describe the motion of a particle with constant acceleration:
vv0at, (2-11)
(2-15)
(2-16)
(2-17)
(2-18)
These are not valid when the acceleration is not constant.
Free-Fall Acceleration An important example of straight-
line motion with constant acceleration is that of an object rising or
falling freely near Earth’s surface. The constant acceleration equa-
tions describe this motion, but we make two changes in notation:
(1) we refer the motion to the vertical yaxis with yvertically up;
(2) we replace awith g, where gis the magnitude of the free-fall
acceleration. Near Earth’s surface, g9.8 m/s2(32 ft/s2).
xx0vt 1
2at2.
xx01
2(v0v)t,
v2v0
22a(xx0),
xx0v0t1
2at2,
adv
dt d2x
dt2.
Questions
1Figure 2-16 gives the velocity of a
particle moving on an xaxis. What
are (a) the initial and (b) the final di-
rections of travel? (c) Does the parti-
cle stop momentarily? (d) Is the ac-
celeration positive or negative? (e) Is
it constant or varying?
2Figure 2-17 gives the accelera-
tion a(t) of a Chihuahua as it chases
a German shepherd along an axis. In
which of the time periods indicated
does the Chihuahua move at constant speed?
is the sign of the particle’s position?
Is the particle’s velocity positive,
negative, or 0 at (b) t1 s, (c) t2
s, and (d) t3 s? (e) How many
times does the particle go through
the point x0?
5Figure 2-20 gives the velocity of
a particle moving along an axis.
Point 1 is at the highest point on the
curve; point 4 is at the lowest point;
and points 2 and 6 are at the same
height. What is the direction of
travel at (a) time t0 and (b) point
4? (c) At which of the six numbered
points does the particle reverse its
direction of travel? (d) Rank the six
points according to the magnitude
of the acceleration, greatest first.
6At t0, a particle moving along an
xaxis is at position x020 m. The
signs of the particle’s initial velocity v0
(at time t0) and constant acceleration a
are, respectively, for four situations: (1)
,; (2) ,; (3) ,; (4) ,.In
which situations will the particle (a)
stop momentarily, (b) pass through the
origin, and (c) never pass through the
origin?
7Hanging over the railing of a
bridge, you drop an egg (no initial ve-
locity) as you throw a second egg
downward. Which curves in Fig. 2-21
a
AB C D E F G H
t
Figure 2-17 Question 2.
t (s)
x
34210
Figure 2-19 Question 4.
t
v
Figure 2-16 Question 1.
v
1
26
35
4
t
Figure 2-20 Question 5.
3Figure 2-18 shows four paths along
which objects move from a starting
point to a final point, all in the same
time interval. The paths pass over a
grid of equally spaced straight lines.
Rank the paths according to (a) the av-
erage velocity of the objects and (b)
the average speed of the objects, great-
est first.
4Figure 2-19 is a graph of a parti-
cle’s position along an xaxis versus time. (a) At time t0, what
3
2
1
4
Figure 2-18 Question 3.
Figure 2-21 Question 7.
0
t
B
A
C
D
EFG
v
apple’s release, the balloon is accelerating upward with a magni-
tude of 4.0 m/s2and has an upward velocity of magnitude 2 m/s.
What are the (a) magnitude and (b) direction of the acceleration of
the apple just after it is released? (c) Just then, is the apple moving
upward or downward, or is it stationary? (d) What is the magni-
tude of its velocity just then? (e) In the next few moments, does the
speed of the apple increase, decrease,or remain constant?
11 Figure 2-23 shows that a particle moving along an xaxis un-
dergoes three periods of acceleration. Without written computa-
tion, rank the acceleration periods according to the increases
they produce in the particle’s velocity, greatest first.
“Cogito ergo zoom!” (I think, therefore I go fast!). In 2001, Sam
Whittingham beat Huber’s record by 19.0 km/h. What was
Whittingham’s time through the 200 m?
••7 Two trains, each having a speed of 30 km/h, are headed at
each other on the same straight track. A bird that can fly 60 km/h
flies off the front of one train when they are 60 km apart and heads
directly for the other train. On reaching the other train, the (crazy)
bird flies directly back to the first train, and so forth.What is the to-
tal distance the bird travels before the trains collide?
••8 Panic escape. Figure 2-24 shows a general situation in
which a stream of people attempt to escape through an exit door
that turns out to be locked. The people move toward the door at
speed vs3.50 m/s, are each d0.25 m in depth, and are sepa-
rated by L1.75 m. The
arrangement in Fig. 2-24
occurs at time t0. (a) At
what average rate does the
layer of people at the door
increase? (b) At what time
does the layer’s depth reach
5.0 m? (The answers reveal
how quickly such a situation
becomes dangerous.)
••9 In 1 km races,runner 1 on track 1 (with time 2 min, 27.95 s)
appears to be faster than runner 2 on track 2 (2 min, 28.15 s).
However, length L2of track 2 might be slightly greater than length
L1of track 1. How large can L2L1be for us still to conclude that
runner 1 is faster?
ILW
Module 2-1 Position, Displacement, and Average Velocity
•1 While driving a car at 90 km/h, how far do you move while
your eyes shut for 0.50 s during a hard sneeze?
•2 Compute your average velocity in the following two cases:
(a) You walk 73.2 m at a speed of 1.22 m/s and then run 73.2 m at a
speed of 3.05 m/s along a straight track. (b) You walk for 1.00 min
at a speed of 1.22 m/s and then run for 1.00 min at 3.05 m/s along a
straight track. (c) Graph xversus tfor both cases and indicate how
the average velocity is found on the graph.
•3 An automobile travels on a straight road for
40 km at 30 km/h. It then continues in the same direction for an-
other 40 km at 60 km/h. (a) What is the average velocity of the car
during the full 80 km trip? (Assume that it moves in the positive x
direction.) (b) What is the average speed? (c) Graph xversus tand
indicate how the average velocity is found on the graph.
•4 A car moves uphill at 40 km/h and then back downhill at 60
km/h.What is the average speed for the round trip?
•5 The position of an object moving along an xaxis is given
by x3t4t2t3, where xis in meters and tin seconds. Find the
position of the object at the following values of t: (a) 1 s, (b) 2 s,
(c) 3 s,and (d) 4 s. (e) What is the object’s displacement between t0
and t4 s? (f) What is its average velocity for the time interval
from t2 s to t4 s? (g) Graph xversus tfor 0 t4 s and indi-
cate how the answer for (f) can be found on the graph.
•6 The 1992 world speed record for a bicycle (human-powered
vehicle) was set by Chris Huber. His time through the measured
200 m stretch was a sizzling 6.509 s, at which he commented,
SSM
WWWSSM
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
Locked
door
L L L
ddd
Figure 2-24 Problem 8.
32 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
give the velocity v(t) for (a) the dropped egg
and (b) the thrown egg? (Curves Aand Bare
parallel; so are C,D, and E; so are Fand G.)
8The following equations give the velocity
v(t) of a particle in four situations: (a) v3; (b)
v4t22t6; (c) v3t4; (d) v5t23.
To which of these situations do the equations of
Table 2-1 apply?
9In Fig. 2-22, a cream tangerine is thrown di-
rectly upward past three evenly spaced windows
of equal heights. Rank the windows according
to (a) the average speed of the cream tangerine
while passing them, (b) the time the cream tan-
gerine takes to pass them, (c) the magnitude of
the acceleration of the cream tangerine while
passing them, and (d) the change vin the
speed of the cream tangerine during the pas-
sage, greatest first.
10 Suppose that a passenger intent on lunch
during his first ride in a hot-air balloon accidently drops an apple
over the side during the balloon’s liftoff. At the moment of the
1
2
3
Figure 2-22
Question 9.
Acceleration a
Time t
(1)
(2)
(3)
Figure 2-23 Question 11.
33
PROBLEMS
••10 To set a speed record in a measured (straight-line)
distance d, a race car must be driven first in one direction (in time t1)
and then in the opposite direction (in time t2). (a) To eliminate the ef-
fects of the wind and obtain the car’s speed vcin a windless situation,
should we find the average of d/t1and d/t2(method 1) or should we di-
vide dby the average of t1and t2? (b) What is the fractional difference
in the two methods when a steady wind blows along the car’s route
and the ratio of the wind speed vwto the car’s speed vcis 0.0240?
••11 You are to drive 300 km to an interview. The interview is
Car Buffer
dL dL L L L
v v
s
Figure 2-25 Problem 12.
•••13 You drive on Interstate 10 from San Antonio to Houston,
half the time at 55 km/h and the other half at 90 km/h. On the way
back you travel half the distance at 55 km/h and the other half at
90 km/h. What is your average speed (a) from San Antonio to
Houston, (b) from Houston back to San Antonio, and (c) for the entire
trip? (d) What is your average velocity for the entire trip? (e) Sketch x
versus tfor (a), assuming the motion is all in the positive xdirec-
tion.Indicate how the average velocity can be found on the sketch.
Module 2-2 Instantaneous Velocity and Speed
•14 An electron moving along the xaxis has a position given
by x16tetm, where tis in seconds. How far is the electron from
the origin when it momentarily stops?
•15 (a) If a particle’s position is given by x412t3t2
(where tis in seconds and xis in meters), what is its velocity at
s? (b) Is it moving in the positive or negative direction of x
just then? (c) What is its speed just then? (d) Is the speed
increasing or decreasing just then? (Try answering the next two
questions without further calculation.) (e) Is there ever an instant
when the velocity is zero? If so, give the time t; if not, answer no.
(f) Is there a time after t3 s when the particle is moving in the
negative direction of x? If so, give the time t; if not, answer no.
•16 The position function x(t) of a particle moving along an xaxis
is x4.0 6.0t2, with xin meters and tin seconds. (a) At what
time and (b) where does the particle (momentarily) stop? At what
(c) negative time and (d) positive time does the particle pass
through the origin? (e) Graph xversus tfor the range 5 s to 5s.
(f) To shift the curve rightward on the graph, should we include the
t1
ILW
term 20tor the term 20tin x(t)? (g) Does that inclusion increase
or decrease the value of xat which the particle momentarily stops?
••17 The position of a particle moving along the xaxis is given in
centimeters by x9.75 1.50t3, where tis in seconds. Calculate (a)
the average velocity during the time interval t2.00 s to t3.00 s;
(b) the instantaneous velocity at t2.00 s; (c) the instantaneous ve-
locity at t3.00 s; (d) the instantaneous velocity at t2.50 s; and
(e) the instantaneous velocity when the particle is midway between
its positions at t2.00 s and t3.00 s. (f) Graph xversus tand in-
dicate your answers graphically.
Module 2-3 Acceleration
•18 The position of a particle moving along an xaxis is given by
x12t22t3, where xis in meters and tis in seconds. Determine (a)
the position,(b) the velocity,and (c) the acceleration of the particle at
t3.0 s. (d) What is the maximum positive coordinate reached by
the particle and (e) at what time is it reached? (f) What is the maxi-
mum positive velocity reached by the particle and (g) at what time is
it reached? (h) What is the acceleration of the particle at the instant
the particle is not moving (other than at t0)? (i) Determine the av-
erage velocity of the particle between t0 and t3s.
•19 At a certain time a particle had a speed of 18 m/s in
the positive xdirection, and 2.4 s later its speed was 30 m/s in the
opposite direction.What is the average acceleration of the particle
during this 2.4 s interval?
•20 (a) If the position of a particle is given by x20t5t3,
where xis in meters and tis in seconds, when, if ever, is the parti-
cle’s velocity zero? (b) When is its acceleration azero? (c) For
what time range (positive or negative) is anegative? (d) Positive?
(e) Graph x(t), v(t), and a(t).
••21 From t0 to t5.00 min, a man stands still, and from
t5.00 min to t10.0 min, he walks briskly in a straight line at a
constant speed of 2.20 m/s. What are (a) his average velocity vavg
and (b) his average acceleration aavg in the time interval 2.00 min to
8.00 min? What are (c) vavg and (d) aavg in the time interval 3.00 min
to 9.00 min? (e) Sketch xversus tand vversus t, and indicate how
the answers to (a) through (d) can be obtained from the graphs.
••22 The position of a particle moving along the xaxis depends on
the time according to the equation xct2bt3, where xis in me-
ters and tin seconds.What are the units of (a) constant cand (b) con-
stant b? Let their numerical values be 3.0 and 2.0,respectively. (c) At
what time does the particle reach its maximum positive xposition?
From t0.0 s to t4.0 s, (d) what distance does the particle move
and (e) what is its displacement? Find its velocity at times (f) 1.0 s,
(g) 2.0 s, (h) 3.0 s, and (i) 4.0 s. Find its acceleration at times (j) 1.0 s,
(k) 2.0 s, (l) 3.0 s,and (m) 4.0 s.
Module 2-4 Constant Acceleration
•23 An electron with an initial velocity v01.50 105m/s
enters a region of length L1.00
cm where it is electrically acceler-
ated (Fig. 2-26). It emerges with
v5.70 106m/s. What is its ac-
celeration, assumed constant?
•24 Catapulting mush-
rooms. Certain mushrooms launch
their spores by a catapult mecha-
nism.As water condenses from the
air onto a spore that is attached to
SSM
SSM
Nonaccelerating
region
Accelerating
region
Path of
electron
L
Figure 2-26 Problem 23.
at 1115 A.M. You plan to drive at 100 km/h, so you leave at 800
A.M. to allow some extra time. You drive at that speed for the first
100 km, but then construction work forces you to slow to 40 km/h
for 40 km.What would be the least speed needed for the rest of the
trip to arrive in time for the interview?
•••12 Traffic shock wave. An abrupt slowdown in concen-
trated traffic can travel as a pulse, termed a shock wave, along the
line of cars, either downstream (in the traffic direction) or up-
stream, or it can be stationary. Figure 2-25 shows a uniformly
spaced line of cars moving at speed v25.0 m/s toward a uni-
formly spaced line of slow cars moving at speed vs5.00 m/s.
Assume that each faster car adds length L12.0 m (car length
plus buffer zone) to the line of slow cars when it joins the line, and as-
sume it slows abruptly at the last instant. (a) For what separation dis-
tance dbetween the faster cars does the shock wave remain
stationary? If the separation is twice that amount, what are the (b)
speed and (c) direction (upstream or downstream) of the shock wave?
34 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
x
Green
car
Red
car xg
xr
Figure 2-27 Problems 34 and 35.
••35 Figure 2-27 shows a red car
and a green car that move toward
each other. Figure 2-28 is a graph of
their motion, showing the positions
xg0270 m and xr035.0 m at
time t0. The green car has a con-
stant speed of 20.0 m/s and the red
car begins from rest. What is the ac-
celeration magnitude of the red car?
••36 A car moves along an xaxis through a distance of 900 m,
starting at rest (at x0) and ending at rest (at x900 m).
Through the first of that distance, its acceleration is 2.25 m/s2.
Through the rest of that distance, its acceleration is 0.750 m/s2.
What are (a) its travel time through the 900 m and (b) its maxi-
mum speed? (c) Graph position x, velocity v, and acceleration a
versus time tfor the trip.
••37 Figure 2-29 depicts the motion
of a particle moving along an xaxis
with a constant acceleration. The fig-
ure’s vertical scaling is set by xs6.0 m.
What are the (a) magnitude and (b) di-
rection of the particle’s acceleration?
••38 (a) If the maximum acceleration
that is tolerable for passengers in a
subway train is 1.34 m/s2and subway
stations are located 806 m apart, what
is the maximum speed a subway train
can attain between stations? (b) What
is the travel time between stations? (c) If a subway train stops for 20 s
at each station, what is the maximum average speed of the train, from
one start-up to the next? (d) Graph x,v, and aversus tfor the interval
from one start-up to the next.
••39 Cars Aand Bmove in
the same direction in adjacent
lanes.The position xof car Ais
given in Fig. 2-30, from time
t0 to t7.0 s. The figure’s
vertical scaling is set by xs
32.0 m.At t0, car Bis at x
0, with a velocity of 12 m/s and
a negative constant accelera-
tion aB. (a) What must aBbe
such that the cars are (momen-
tarily) side by side (momentarily at the same value of x) at t4.0 s?
(b) For that value of aB, how many times are the cars side by side?
(c) Sketch the position xof car Bversus time ton Fig. 2-30. How
many times will the cars be side by side if the magnitude of accelera-
tion aBis (d) more than and (e) less than the answer to part (a)?
••40 You are driving toward a traffic signal when it turns yel-
low. Your speed is the legal speed limit of v055 km/h; your best
deceleration rate has the magnitude a5.18 m/s2.Your best reaction
time to begin braking is T0.75 s. To avoid having the front of your
car enter the intersection after the light turns red, should you
brake to a stop or continue to move at 55 km/h if the distance to
1
4
the mushroom, a drop grows on one side of the spore and a film
grows on the other side.The spore is bent over by the drop’s weight,
but when the film reaches the drop, the drop’s water suddenly
spreads into the film and the spore springs upward so rapidly that it
is slung off into the air. Typically, the spore reaches a speed of 1.6
m/s in a 5.0 mm launch; its speed is then reduced to zero in 1.0 mm
by the air. Using those data and assuming constant accelerations,
find the acceleration in terms of gduring (a) the launch and (b) the
speed reduction.
•25 An electric vehicle starts from rest and accelerates at a rate
of 2.0 m/s2in a straight line until it reaches a speed of 20 m/s. The
vehicle then slows at a constant rate of 1.0 m/s2until it stops. (a)
How much time elapses from start to stop? (b) How far does the
vehicle travel from start to stop?
•26 A muon (an elementary particle) enters a region with a speed
of 5.00 106m/s and then is slowed at the rate of 1.25 1014 m/s2.
(a) How far does the muon take to stop? (b) Graph xversus tand v
versus tfor the muon.
•27 An electron has a constant acceleration of 3.2 m/s2. At a
certain instant its velocity is 9.6 m/s. What is its velocity (a) 2.5 s
earlier and (b) 2.5 s later?
•28 On a dry road, a car with good tires may be able to brake
with a constant deceleration of 4.92 m/s2. (a) How long does such
a car, initially traveling at 24.6 m/s, take to stop? (b) How far does
it travel in this time? (c) Graph xversus tand vversus tfor the
deceleration.
•29 A certain elevator cab has a total run of 190 m and a max-
imum speed of 305 m/min, and it accelerates from rest and then
back to rest at 1.22 m/s2. (a) How far does the cab move while ac-
celerating to full speed from rest? (b) How long does it take to
make the nonstop 190 m run, starting and ending at rest?
•30 The brakes on your car can slow you at a rate of 5.2 m/s2. (a)
If you are going 137 km/h and suddenly see a state trooper, what is
the minimum time in which you can get your car under the 90 km/h
speed limit? (The answer reveals the futility of braking to keep
your high speed from being detected with a radar or laser gun.)
(b) Graph xversus tand vversus tfor such a slowing.
•31 Suppose a rocket ship in deep space moves with con-
stant acceleration equal to 9.8 m/s2, which gives the illusion of nor-
mal gravity during the flight. (a) If it starts from rest, how long will
it take to acquire a speed one-tenth that of light, which travels at
3.0 108m/s? (b) How far will it travel in so doing?
•32 A world’s land speed record was set by Colonel John
P. Stapp when in March 1954 he rode a rocket-propelled sled that
moved along a track at 1020 km/h. He and the sled were brought to
a stop in 1.4 s. (See Fig. 2-7.) In terms of g, what acceleration did he
experience while stopping?
•33 A car traveling 56.0 km/h is 24.0 m from a barrier
when the driver slams on the brakes. The car hits the barrier 2.00 s
later. (a) What is the magnitude of the car’s constant acceleration
before impact? (b) How fast is the car traveling at impact?
••34 In Fig. 2-27, a red car and a green car,identical except for the
color, move toward each other in adjacent lanes and parallel to an x
axis. At time t0, the red car is at xr0 and the green car is at xg
220 m. If the red car has a constant velocity of 20 km/h, the cars pass
each other at x44.5 m, and if it has a constant velocity of 40 km/h,
they pass each other at x76.6 m. What are (a) the initial velocity
and (b) the constant acceleration of the green car?
ILWSSM
SSM
ILW
xg0
xr0
0
x (m)
t (s)
120
Figure 2-28 Problem 35.
12
0
x(m)
xs
t(s)
Figure 2-29 Problem 37.
x(m)
xs
0123
t (s)
4567
Figure 2-30 Problem 39.
35
PROBLEMS
the intersection and the duration of the yellow light are (a) 40 m and
2.8 s, and (b) 32 m and 1.8 s? Give an answer of brake,continue, either
(if either strategy works), or neither (if neither strategy works and the
yellow duration is inappropriate).
••41 As two trains move
along a track, their conductors
suddenly notice that they are
headed toward each other.
Figure 2-31 gives their velocities
vas functions of time tas the
conductors slow the trains. The
figure’s vertical scaling is set by
vs40.0 m/s. The slowing
processes begin when the trains are 200 m apart.What is their separa-
tion when both trains have stopped?
•••42 You are arguing over a cell phone while trailing an
unmarked police car by 25 m; both your car and the police car are
traveling at 110 km/h. Your argument diverts your attention from
the police car for 2.0 s (long enough for you to look at the phone
and yell, “I won’t do that!”). At the beginning of that 2.0 s, the po-
lice officer begins braking suddenly at 5.0 m/s2. (a) What is the sep-
aration between the two cars when your attention finally returns?
Suppose that you take another 0.40 s to realize your danger and
begin braking. (b) If you too brake at 5.0 m/s2, what is your speed
when you hit the police car?
•••43 When a high-speed passenger train traveling at
161 km/h rounds a bend, the engineer is shocked to see that a
locomotive has
improperly entered onto the track from a siding
and is a distance D
676 m ahead (Fig. 2-32). The locomotive is
moving at 29.0 km/h. The engineer of the high-speed train imme-
diately applies the brakes. (a) What must be the magnitude of the
resulting constant deceleration if a collision is to be just avoided?
(b) Assume that the engineer is at x0 when, at t0, he first
•46 Raindrops fall 1700 m from a cloud to the ground. (a) If they
were not slowed by air resistance, how fast would the drops be
moving when they struck the ground? (b) Would it be safe to walk
outside during a rainstorm?
•47 At a construction site a pipe wrench struck the ground
with a speed of 24 m/s. (a) From what height was it inadvertently
dropped? (b) How long was it falling? (c) Sketch graphs of y,v,
and aversus tfor the wrench.
•48 A hoodlum throws a stone vertically downward with an ini-
tial speed of 12.0 m/s from the roof of a building, 30.0 m above the
ground. (a) How long does it take the stone to reach the ground?
(b) What is the speed of the stone at impact?
•49 A hot-air balloon is ascending at the rate of 12 m/s and
is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take to reach the ground? (b) With
what speed does it hit the ground?
••50 At time t0, apple 1 is dropped from a bridge onto a road-
way beneath the bridge; somewhat later, apple 2 is thrown down
from the same height. Figure 2-33 gives the vertical positions yof
the apples versus tduring the falling, until both apples have hit the
roadway. The scaling is set by ts2.0 s. With approximately what
speed is apple 2 thrown down?
SSM
SSM
v (m/s)
0
vs
t (s)
246
Figure 2-31 Problem 41.
D
High-speed
train Locomotive
Figure 2-32 Problem 43.
y
0ts
Figure 2-33 Problem 50.
••51 As a runaway scientific bal-
loon ascends at 19.6 m/s, one of its
instrument packages breaks free of a
harness and free-falls. Figure 2-34
gives the vertical velocity of the
package versus time, from before it
breaks free to when it reaches the
ground. (a) What maximum height
above the break-free point does it
rise? (b) How high is the break-free
point above the ground?
••52 A bolt is dropped from a bridge under construction,
falling 90 m to the valley below the bridge. (a) In how much
time does it pass through the last 20% of its fall? What is its speed
(b) when it begins that last 20% of its fall and (c) when it reaches
the valley beneath the bridge?
••53 A key falls from a bridge that is 45 m above the
water. It falls directly into a model boat, moving with constant
velocity, that is 12 m from the point of impact when the key is re-
leased.What is the speed of the boat?
••54 A stone is dropped into a river from a bridge 43.9 m
above the water. Another stone is thrown vertically down 1.00 s
after the first is dropped. The stones strike the water at the same
time. (a) What is the initial speed of the second stone? (b) Plot
velocity versus time on a graph for each stone, taking zero time as
the instant the first stone is released.
ILWSSM
Figure 2-34 Problem 51.
t(s)
v
2
0468
Module 2-5 Free-Fall Acceleration
•44 When startled, an armadillo will leap upward. Suppose it
rises 0.544 m in the first 0.200 s. (a) What is its initial speed as it
leaves the ground? (b) What is its speed at the height of 0.544 m?
(c) How much higher does it go?
•45 (a) With what speed must a ball be thrown verti-
cally from ground level to rise to a maximum height of 50 m?
(b) How long will it be in the air? (c) Sketch graphs of y,v, and a
versus tfor the ball. On the first two graphs, indicate the time at
which 50 m is reached.
WWWSSM
spots the locomotive. Sketch x(t) curves for the locomotive and
high-speed train for the cases in which a collision is just avoided
and is not quite avoided.
time t0.At t1.5 s it passes the top of a tall tower, and 1.0 s later
it reaches its maximum height.What is the height of the tower?
•••61 A steel ball is dropped from a building’s roof and passes
a window, taking 0.125 s to fall from the top to the bottom of the
window, a distance of 1.20 m. It then falls to a sidewalk and
bounces back past the window, moving from bottom to top in
0.125 s. Assume that the upward flight is an exact reverse of the
fall. The time the ball spends below the bottom of the window is
2.00 s. How tall is the building?
•••62 A basketball player grabbing a rebound jumps
76.0 cm vertically. How much total time (ascent and descent) does
the player spend (a) in the top 15.0 cm of this jump and (b) in the
bottom 15.0 cm? (The player seems to hang in the air at the top.)
•••63 A drowsy cat spots a flowerpot that sails first up and then
down past an open window.The pot is in view for a total of 0.50 s, and
the top-to-bottom height of the window is 2.00 m. How high above the
window top does the flowerpot go?
•••64 A ball is shot vertically up-
ward from the surface of another
planet. A plot of yversus tfor the
ball is shown in Fig. 2-36, where yis
the height of the ball above its start-
ing point and t0 at the instant the
ball is shot. The figure’s vertical scal-
ing is set by ys30.0 m.What are the
magnitudes of (a) the free-fall accel-
eration on the planet and (b) the ini-
tial velocity of the ball?
36 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
t (ms)
0 50 100 140
vs
v (m/s)
Figure 2-37 Problem 66.
••55 A ball of moist clay falls 15.0 m to the ground. It is
in contact with the ground for 20.0 ms before stopping. (a) What is
the magnitude of the average acceleration of the ball during the time
it is in contact with the ground? (Treat the ball as a particle.) (b) Is the
average acceleration up or down?
••56 Figure 2-35
shows the speed vversus
height yof a ball tossed
directly upward, along a y
axis. Distance dis 0.40 m.
The speed at height yAis
vA.The speed at height yB
is vA. What is speed vA?
••57 To test the quality
of a tennis ball, you drop
it onto the floor from a
height of 4.00 m. It re-
bounds to a height of 2.00 m. If the ball is in contact with the floor
for 12.0 ms, (a) what is the magnitude of its average acceleration
during that contact and (b) is the average acceleration up or down?
••58 An object falls a distance hfrom rest. If it travels 0.50hin
the last 1.00 s, find (a) the time and (b) the height of its fall. (c)
Explain the physically unacceptable solution of the quadratic
equation in tthat you obtain.
••59 Water drips from the nozzle of a shower onto the floor 200
cm below. The drops fall at regular (equal) intervals of time, the
first drop striking the floor at the instant the fourth drop begins to
fall.When the first drop strikes the floor, how far below the nozzle
are the (a) second and (b) third drops?
••60 A rock is thrown vertically upward from ground level at
1
3
SSM Module 2-6 Graphical Integration in Motion Analysis
•65 Figure 2-15agives the acceleration of a volunteer’s
head and torso during a rear-end collision. At maximum head ac-
celeration, what is the speed of (a) the head and (b) the torso?
••66 In a forward punch in karate, the fist begins at rest at
the waist and is brought rapidly forward until the arm is fully ex-
tended. The speed v(t) of the fist is given in Fig. 2-37 for someone
skilled in karate. The vertical scaling is set by vs8.0 m/s. How far
has the fist moved at (a) time t50 ms and (b) when the speed of
the fist is maximum?
••67 When a soccer
ball is kicked to-
ward a player and
the player deflects
the ball by “head-
ing” it, the accelera-
tion of the head dur-
ing the collision can
be significant. Figure
2-38 gives the meas-
ured acceleration
a(t) of a soccer player’s head for a bare head and a helmeted head,
starting from rest. The scaling on the vertical axis is set by as200
m/s2.At time t7.0 ms, what is the difference in the speed acquired
by the bare head and the speed acquired by the helmeted head?
••68 A salamander of the genus Hydromantes captures
prey by launching its tongue
as a projectile: The skeletal
part of the tongue is shot for-
ward, unfolding the rest of
the tongue, until the outer
portion lands on the prey,
sticking to it. Figure 2-39
shows the acceleration mag-
nitude aversus time tfor the
acceleration phase of the
launch in a typical situation.
The indicated accelerations are
a2400 m/s2and a1100 m/s2.
What is the outward speed of the
tongue at the end of the
acceleration phase?
••69 How far does the run-
ner whose velocity–time graph is
shown in Fig. 2-40 travel in 16 s?
The figure’s vertical scaling is set
by vs8.0 m/s.
ILW
y
yAyB
vA
vA
v
1
_
_
3
d
0
Figure 2-35 Problem 56.
ys
0012345
y(m)
t(s)
Figure 2-36 Problem 64.
a(m/s2)
a2
a1
0
t (ms)
10 20 30 40
Figure 2-39 Problem 68.
0481216
v
s
v(m/s)
t(s)
Figure 2-40 Problem 69.
Bare
Helmet
0
a
s
246
a (m/s2)
t (ms)
Figure 2-38 Problem 67.
37
PROBLEMS
•••70 Two particles move along an xaxis. The position of particle 1
is given by x6.00t
2
3.00t2.00 (in meters and seconds); the ac-
celeration of particle 2 is given by a8.00t(in meters per second
squared and seconds) and, at t0, its velocity is 20 m/s. When the
velocities of the particles match, what is their velocity?
Additional Problems
71 In an arcade video game, a spot is programmed to move
across the screen according to x9.00t0.750t3, where xis dis-
tance in centimeters measured from the left edge of the screen and
tis time in seconds. When the spot reaches a screen edge, at either
x0 or x15.0 cm, tis reset to 0 and the spot starts moving again
according to x(t). (a) At what time after starting is the spot instan-
taneously at rest? (b) At what value of xdoes this occur? (c) What
is the spot’s acceleration (including sign) when this occurs? (d)
Is it moving right or left just prior to coming to rest? (e) Just after?
(f) At what time t0 does it first reach an edge of the screen?
72 A rock is shot vertically upward from the edge of the top of a
tall building.The rock reaches its maximum height above the top of
the building 1.60 s after being shot. Then, after barely missing the
edge of the building as it falls downward, the rock strikes the ground
6.00 s after it is launched. In SI units: (a) with what upward velocity
is the rock shot, (b) what maximum height above the top of the
building is reached by the rock, and (c) how tall is the building?
73 At the instant the traffic light turns green, an automobile
starts with a constant acceleration aof 2.2 m/s2.At the same instant
a truck, traveling with a constant speed of 9.5 m/s, overtakes and
passes the automobile. (a) How far beyond the traffic signal will
the automobile overtake the truck? (b) How fast will the automo-
bile be traveling at that instant?
74 A pilot flies horizontally at 1300 km/h, at height h35 m
above initially level ground. However, at time t0, the pilot be-
gins to fly over ground sloping upward at angle u4.3° (Fig. 2-41).
If the pilot does not change the airplane’s heading, at what time t
does the plane strike the ground?
θ
h
Figure 2-41 Problem 74.
75 To stop a car, first you require a certain reaction time to be-
gin braking; then the car slows at a constant rate. Suppose that the
total distance moved by your car during these two phases is 56.7 m
when its initial speed is 80.5 km/h, and 24.4 m when its initial speed
is 48.3 km/h. What are (a) your reaction time and (b) the magni-
tude of the acceleration?
76 Figure 2-42 shows part of a street where traffic flow
is to be controlled to allow a platoon of cars to move smoothly
along the street. Suppose that the platoon leaders have just
ONE WAY
3
2
1
D23
D12
Figure 2-42 Problem 76.
reached intersection 2, where the green appeared when they were
distance dfrom the intersection. They continue to travel at a cer-
tain speed vp(the speed limit) to reach intersection 3, where the
green appears when they are distance dfrom it. The intersections
are separated by distances D23 and D12. (a) What should be the
time delay of the onset of green at intersection 3 relative to that at
intersection 2 to keep the platoon moving smoothly?
Suppose, instead, that the platoon had been stopped by a red
light at intersection 1. When the green comes on there, the leaders
require a certain time trto respond to the change and an additional
time to accelerate at some rate ato the cruising speed vp. (b) If the
green at intersection 2 is to appear when the leaders are distance d
from that intersection, how long after the light at intersection 1
turns green should the light at intersection 2 turn green?
77 A hot rod can accelerate from 0 to 60 km/h in 5.4 s.
(a) What is its average acceleration, in m/s2, during this time? (b)
How far will it travel during the 5.4 s, assuming its acceleration is con-
stant? (c) From rest, how much time would it require to go a distance
of 0.25 km if its acceleration could be maintained at the value in (a)?
78 A red train traveling at 72 km/h and a green train traveling
at 144 km/h are headed toward each other along a straight, level
track. When they are 950 m apart, each engineer sees the other’s
train and applies the brakes. The brakes slow each train at the rate
of 1.0 m/s2. Is there a collision? If so, answer yes and give the speed
of the red train and the speed of the green train at impact, respec-
tively. If not, answer no and give the separation between the trains
when they stop.
79 At time t0, a rock
climber accidentally allows a
piton to fall freely from a high
point on the rock wall to the
valley below him.Then, after a
short delay, his climbing part-
ner, who is 10 m higher on the
wall, throws a piton down-
ward. The positions yof the
pitons versus tduring the
falling are given in Fig. 2-43.
With what speed is the second piton thrown?
80 A train started from rest and moved with constant accelera-
tion. At one time it was traveling 30 m/s, and 160 m farther on it
was traveling 50 m/s. Calculate (a) the acceleration, (b) the time re-
quired to travel the 160 m mentioned, (c) the time required to at-
tain the speed of 30 m/s, and (d) the distance moved from rest to
the time the train had a speed of 30 m/s. (e) Graph xversus tand v
versus tfor the train, from rest.
81 A particle’s acceleration along an xaxis is a5.0t, with t
in seconds and ain meters per
second squared. At t2.0 s,
its velocity is 17 m/s.What is
its velocity at t4.0 s?
82 Figure 2-44 gives the ac-
celeration aversus time tfor
a particle moving along an x
axis.The a-axis scale is set by
as12.0 m/s2. At t2.0 s,
the particle’s velocity is 7.0
m/s.What is its velocity at t
6.0 s?
SSM
SSM
y
t (s)
0123
Figure 2-43 Problem 79.
t(s)
–2 0 246
as
a (m/s2)
Figure 2-44 Problem 82.
38 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
83 Figure 2-45 shows a simple device for measuring your
reaction time. It consists of a cardboard strip marked with a scale
and two large dots. A friend holds the strip vertically, with thumb
and forefinger at the dot on the right in Fig. 2-45. You then posi-
tion your thumb and forefinger at the other dot (on the left in
Fig. 2-45), being careful not to touch the strip. Your friend re-
leases the strip, and you try to pinch it as soon as possible after
you see it begin to fall.The mark at the place where you pinch the
strip gives your reaction time. (a) How far from the lower dot
should you place the 50.0 ms mark? How much higher should
you place the marks for (b) 100, (c) 150, (d) 200, and (e) 250 ms?
(For example, should the 100 ms marker be 2 times as far from
the dot as the 50 ms marker? If so, give an answer of 2 times. Can
you find any pattern in the answers?)
the acceleration of the particle at t5.0 s? (d) What is the average ve-
locity of the particle between t1.0 s and t5.0 s? (e) What is the
average acceleration of the particle between t1.0 s and t5.0 s?
91 A rock is dropped from a 100-m-high cliff. How long does it
take to fall (a) the first 50 m and (b) the second 50 m?
92 Two subway stops are separated by 1100 m. If a subway train
accelerates at 1.2 m/s2from rest through the first half of the dis-
tance and decelerates at 1.2 m/s2through the second half, what
are (a) its travel time and (b) its maximum speed? (c) Graph x,v,
and aversus tfor the trip.
93 A stone is thrown vertically upward. On its way up it passes
point Awith speed v, and point B, 3.00 m higher than A, with speed
Calculate (a) the speed vand (b) the maximum height reached
by the stone above point B.
94 A rock is dropped (from rest) from the top of a 60-m-tall
building. How far above the ground is the rock 1.2 s before it
reaches the ground?
95 An iceboat has a constant velocity toward the east when
a sudden gust of wind causes the iceboat to have a constant accel-
eration toward the east for a period of 3.0 s. A plot of xversus tis
shown in Fig. 2-47, where t0 is taken to be the instant the wind
starts to blow and the positive xaxis is toward the east. (a) What is
the acceleration of the iceboat during the 3.0 s interval? (b) What
is the velocity of the iceboat at the end of the 3.0 s interval? (c) If
the acceleration remains constant for an additional 3.0 s, how far
does the iceboat travel during this second 3.0 s interval?
SSM
1
2v.
250
200
150
100
50
0
Reaction time (ms)
Figure 2-45 Problem 83.
84 A rocket-driven sled running on a straight, level track is
used to investigate the effects of large accelerations on humans.
One such sled can attain a speed of 1600 km/h in 1.8 s, starting
from rest. Find (a) the acceleration (assumed constant) in terms of
gand (b) the distance traveled.
85 A mining cart is pulled up a hill at 20 km/h and then pulled
back down the hill at 35 km/h through its original level. (The time
required for the cart’s reversal at the top of its climb is negligible.)
What is the average speed of the cart for its round trip, from its
original level back to its original level?
86 A motorcyclist who is moving along an xaxis directed to-
ward the east has an acceleration given by a(6.1 1.2t) m/s2
for 0 t6.0 s. At t0, the velocity and position of the cyclist
are 2.7 m/s and 7.3 m. (a) What is the maximum speed achieved
by the cyclist? (b) What total distance does the cyclist travel be-
tween t0 and 6.0 s?
87 When the legal speed limit for the New York Thruway
was increased from 55 mi/h to 65 mi/h, how much time was saved
by a motorist who drove the 700 km between the Buffalo entrance
and the New York City exit at the legal speed limit?
88 A car moving with constant acceleration covered the distance
between two points 60.0 m apart in 6.00 s. Its speed as it passed the
second point was 15.0 m/s. (a) What was the speed at the first
point? (b) What was the magnitude of the acceleration? (c) At
what prior distance from the first point was the car at rest? (d) Graph
xversus tand vversus tfor the car,from rest (t0).
89 A certain juggler usually tosses balls vertically to
a height H.To what height must they be tossed if they are to spend
twice as much time in the air?
90 A particle starts from the ori-
gin at t0 and moves along the
positive xaxis. A graph of the veloc-
ity of the particle as a function of the
time is shown in Fig. 2-46; the v-axis
scale is set by vs4.0 m/s. (a) What
is the coordinate of the particle at
t5.0 s? (b) What is the velocity of
the particle at t5.0 s? (c) What is
SSM
SSM
Figure 2-47 Problem 95.
30
25
20
15
10
5
00 0.5 1 1.5 2 2.5 3
x(m)
t(s)
v
s
021
t(s)
3456
v(m/s)
Figure 2-46 Problem 90.
96 A lead ball is dropped in a lake from a diving board 5.20 m
above the water. It hits the water with a certain velocity and then
sinks to the bottom with this same constant velocity. It reaches the
bottom 4.80 s after it is dropped. (a) How deep is the lake? What
are the (b) magnitude and (c) direction (up or down) of the aver-
age velocity of the ball for the entire fall? Suppose that all the wa-
ter is drained from the lake.The ball is now thrown from the diving
board so that it again reaches the bottom in 4.80 s. What are the
(d) magnitude and (e) direction of the initial velocity of the ball?
97 The single cable supporting an unoccupied construction ele-
vator breaks when the elevator is at rest at the top of a 120-m-high
building. (a) With what speed does the elevator strike the ground?
(b) How long is it falling? (c) What is its speed when it passes the
halfway point on the way down? (d) How long has it been falling
when it passes the halfway point?
98 Two diamonds begin a free fall from rest from the same
height, 1.0 s apart. How long after the first diamond begins to fall
will the two diamonds be 10 m apart?
99 A ball is thrown vertically downward from the top of a 36.6-
m-tall building. The ball passes the top of a window that is 12.2 m
above the ground 2.00 s after being thrown. What is the speed of
the ball as it passes the top of the window?
39
100 A parachutist bails out and freely falls 50 m. Then the para-
chute opens, and thereafter she decelerates at 2.0 m/s2. She reaches
the ground with a speed of 3.0 m/s. (a) How long is the parachutist
in the air? (b) At what height does the fall begin?
101 A ball is thrown down vertically with an initial speed of v0
from a height of h. (a) What is its speed just before it strikes the
ground? (b) How long does the ball take to reach the ground?
What would be the answers to (c) part a and (d) part b if the ball
were thrown upward from the same height and with the same ini-
tial speed? Before solving any equations, decide whether the an-
swers to (c) and (d) should be greater than, less than, or the same
as in (a) and (b).
102 The sport with the fastest moving ball is jai alai, where
measured speeds have reached 303 km/h. If a professional jai alai
player faces a ball at that speed and involuntarily blinks, he
blacks out the scene for 100 ms. How far does the ball move dur-
ing the blackout?
103 If a baseball pitcher throws a fastball at a horizontal speed of
160 km/h, how long does the ball take to reach home plate 18.4 m
away?
104 A proton moves along the xaxis according to the equation
, where xis in meters and tis in seconds. Calculate (a)
the average velocity of the proton during the first 3.0 s of its motion,
(b) the instantaneous velocity of the proton at t3.0 s, and (c) the
instantaneous acceleration of the proton at t3.0 s. (d) Graph x
versus tand indicate how the answer to (a) can be obtained from the
plot. (e) Indicate the answer to (b) on the graph. (f) Plot vversus t
and indicate on it the answer to (c).
105 A motorcycle is moving at 30 m/s when the rider applies the
brakes, giving the motorcycle a constant deceleration. During the 3.0 s
interval immediately after braking begins, the speed decreases to
15 m/s. What distance does the motorcycle travel from the instant
braking begins until the motorcycle stops?
106 A shuffleboard disk is accelerated at a constant rate from rest
to a speed of 6.0 m/s over a 1.8 m distance by a player using a cue.At
this point the disk loses contact with the cue and slows at a constant
rate of 2.5 m/s2until it stops. (a) How much time elapses from when
the disk begins to accelerate until it stops? (b) What total distance
does the disk travel?
107 The head of a rattlesnake can accelerate at 50 m/s2in striking
a victim. If a car could do as well, how long would it take to reach a
speed of 100 km/h from rest?
108 A jumbo jet must reach a speed of 360 km/h on the runway
for takeoff. What is the lowest constant acceleration needed for
takeoff from a 1.80 km runway?
109 An automobile driver increases the speed at a constant rate
from 25 km/h to 55 km/h in 0.50 min. A bicycle rider speeds up at a
constant rate from rest to 30 km/h in 0.50 min. What are the magni-
tudes of (a) the driver’s acceleration and (b) the rider’s acceleration?
110 On average, an eye blink lasts about 100 ms. How far does a
MiG-25 “Foxbat” fighter travel during a pilot’s blink if the plane’s
average velocity is 3400 km/h?
111 A certain sprinter has a top speed of 11.0 m/s. If the sprinter
starts from rest and accelerates at a constant rate, he is able to
reach his top speed in a distance of 12.0 m. He is then able to main-
tain this top speed for the remainder of a 100 m race. (a) What is
his time for the 100 m race? (b) In order to improve his time, the
sprinter tries to decrease the distance required for him to reach his
x50t10t2
top speed. What must this distance be if he is to achieve a time of
10.0 s for the race?
112 The speed of a bullet is measured to be 640 m/s as the bullet
emerges from a barrel of length 1.20 m.Assuming constant accelera-
tion, find the time that the bullet spends in the barrel after it is fired.
113 The Zero Gravity Research Facility at the NASA Glenn
Research Center includes a 145 m drop tower.This is an evacuated ver-
tical tower through which, among other possibilities, a 1-m-diameter
sphere containing an experimental package can be dropped. (a)
How long is the sphere in free fall? (b) What is its speed just as it
reaches a catching device at the bottom of the tower? (c) When
caught, the sphere experiences an average deceleration of 25gas its
speed is reduced to zero.Through what distance does it travel during
the deceleration?
114 A car can be braked to a stop from the autobahn-like
speed of 200 km/h in 170 m. Assuming the acceleration is constant,
find its magnitude in (a) SI units and (b) in terms of g. (c) How much
time Tbis required for the braking? Your reaction time Tris the time
you require to perceive an emergency, move your foot to the brake,
and begin the braking. If Tr400 ms, then (d) what is Tbin terms of
Tr, and (e) is most of the full time required to stop spent in reacting
or braking? Dark sunglasses delay the visual signals sent from the
eyes to the visual cortex in the brain, increasing Tr.(f) In the extreme
case in which Tris increased by 100 ms, how much farther does
the car travel during your reaction time?
115 In 1889, at Jubbulpore, India, a tug-of-war was finally won af-
ter 2 h 41 min, with the winning team displacing the center of the
rope 3.7 m. In centimeters per minute, what was the magnitude of
the average velocity of that center point during the contest?
116 Most important in an investigation of an airplane crash by the
U.S. National Transportation Safety Board is the data stored on the
airplane’s flight-data recorder, commonly called the “black box” in
spite of its orange coloring and reflective tape. The recorder is engi-
neered to withstand a crash with an average deceleration of magni-
tude 3400gduring a time interval of 6.50 ms. In such a crash, if the
recorder and airplane have zero speed at the end of that time inter-
val, what is their speed at the beginning of the interval?
117 From January 26, 1977, to September 18, 1983, George
Meegan of Great Britain walked from Ushuaia, at the southern tip
of South America, to Prudhoe Bay in Alaska, covering 30 600 km. In
meters per second, what was the magnitude of his average velocity
during that time period?
118 The wings on a stonefly do not flap, and thus the insect cannot
fly. However, when the insect is on a water surface, it can sail across
the surface by lifting its wings into a breeze. Suppose that you time
stoneflies as they move at constant speed along a straight path of a
certain length. On average, the trips each take 7.1 s with the wings
set as sails and 25.0 s with the wings tucked in. (a) What is the ratio of
the sailing speed vsto the nonsailing speed vns? (b) In terms of vs,
what is the difference in the times the insects take to travel the first
2.0 m along the path with and without sailing?
119 The position of a particle as it moves along a yaxis is given by
y(2.0 cm) sin (pt/4),
with tin seconds and yin centimeters. (a) What is the average veloc-
ity of the particle between t0 and t2.0 s? (b) What is the instan-
taneous velocity of the particle at t0, 1.0, and 2.0 s? (c) What is the
average acceleration of the particle between t0 and t2.0 s?
(d) What is the instantaneous acceleration of the particle at t0,
1.0, and 2.0 s?
PROBLEMS
40
CHAPTER 3
Vectors
3-1 VECTORS AND THEIR COMPONENTS
3.01 Add vectors by drawing them in head-to-tail arrange-
ments, applying the commutative and associative laws.
3.02 Subtract a vector from a second one.
3.03 Calculate the components of a vector on a given coordi-
nate system, showing them in a drawing.
3.04 Given the components of a vector, draw the vector
and determine its magnitude and orientation.
3.05 Convert angle measures between degrees and radians.
●Scalars, such as temperature, have magnitude only. They
are specified by a number with a unit (10°C) and obey the
rules of arithmetic and ordinary algebra. Vectors, such as dis-
placement, have both magnitude and direction (5 m, north)
and obey the rules of vector algebra.
●Two vectors and may be added geometrically by draw-
ing them to a common scale and placing them head to tail.
The vector connecting the tail of the first to the head of the
second is the vector sum . To subtract from , reverse the
direction of to get ; then add to . Vector addition is
commutative and obeys the associative law.
a
:
b
:
b
:
b
:a
:
b
:
s
:
b
:
a
:
●The (scalar) components and of any two-dimensional
vector along the coordinate axes are found by dropping
perpendicular lines from the ends of onto the coordinate
axes. The components are given by
axacos uand ayasin u,
where uis the angle between the positive direction of the x
axis and the direction of . The algebraic sign of a component
indicates its direction along the associated axis. Given its
components, we can find the magnitude and orientation of
the vector with
and .tan
ay
ax
a2a2
xa2
y
a
:
a
:
a
:
a
:
ay
ax
What Is Physics?
Physics deals with a great many quantities that have both size and direction, and it
needs a special mathematical language—the language of vectors—to describe
those quantities. This language is also used in engineering, the other sciences, and
even in common speech. If you have ever given directions such as “Go five blocks
down this street and then hang a left,” you have used the language of vectors. In
fact, navigation of any sort is based on vectors, but physics and engineering also
need vectors in special ways to explain phenomena involving rotation and mag-
netic forces, which we get to in later chapters. In this chapter, we focus on the basic
language of vectors.
Vectors and Scalars
A particle moving along a straight line can move in only two directions. We can
take its motion to be positive in one of these directions and negative in the other.
For a particle moving in three dimensions, however,a plus sign or minus sign is no
longer enough to indicate a direction. Instead, we must use a vector.
Key Ideas
Learning Objectives
After reading this module, you should be able to . . .
3-1 VECTORS AND THEIR COMPONENTS 41
Avector has magnitude as well as direction, and vectors follow certain
(vector) rules of combination, which we examine in this chapter. A vector
quantity is a quantity that has both a magnitude and a direction and thus can be
represented with a vector. Some physical quantities that are vector quantities are
displacement, velocity, and acceleration. You will see many more throughout this
book, so learning the rules of vector combination now will help you greatly in
later chapters.
Not all physical quantities involve a direction.Temperature, pressure, energy,
mass, and time, for example, do not “point” in the spatial sense. We call such
quantities scalars, and we deal with them by the rules of ordinary algebra.A sin-
gle value, with a sign (as in a temperature of 40°F), specifies a scalar.
The simplest vector quantity is displacement, or change of position. A vec-
tor that represents a displacement is called, reasonably, a displacement vector.
(Similarly, we have velocity vectors and acceleration vectors.) If a particle changes
its position by moving from Ato Bin Fig.3-1a, we say that it undergoes a displace-
ment from Ato B, which we represent with an arrow pointing from Ato B.The ar-
row specifies the vector graphically. To distinguish vector symbols from other
kinds of arrows in this book, we use the outline of a triangle as the arrowhead.
In Fig. 3-1a, the arrows from Ato B, from Ato B, and from Ato Bhave
the same magnitude and direction.Thus, they specify identical displacement vec-
tors and represent the same change of position for the particle. A vector can be
shifted without changing its value if its length and direction are not changed.
The displacement vector tells us nothing about the actual path that the parti-
cle takes. In Fig. 3-1b, for example, all three paths connecting points Aand Bcor-
respond to the same displacement vector, that of Fig. 3-1a. Displacement vectors
represent only the overall effect of the motion, not the motion itself.
Adding Vectors Geometrically
Suppose that, as in the vector diagram of Fig. 3-2a, a particle moves from Ato B
and then later from Bto C.We can represent its overall displacement (no matter
what its actual path) with two successive displacement vectors, AB and BC.
The net displacement of these two displacements is a single displacement from A
to C. We call AC the vector sum (or resultant) of the vectors AB and BC. This
sum is not the usual algebraic sum.
In Fig. 3-2b, we redraw the vectors of Fig. 3-2aand relabel them in the way
that we shall use from now on, namely, with an arrow over an italic symbol, as
in .If we want to indicate only the magnitude of the vector (a quantity that lacks
a sign or direction), we shall use the italic symbol, as in a,b, and s. (You can use
just a handwritten symbol.) A symbol with an overhead arrow always implies
both properties of a vector,magnitude and direction.
We can represent the relation among the three vectors in Fig. 3-2bwith the
vector equation
(3-1)
which says that the vector is the vector sum of vectors and .The symbol in
Eq. 3-1 and the words “sum” and “add” have different meanings for vectors than
they do in the usual algebra because they involve both magnitude and direction.
Figure 3-2 suggests a procedure for adding two-dimensional vectors and
geometrically. (1) On paper, sketch vector to some convenient scale and at the
proper angle. (2) Sketch vector to the same scale, with its tail at the head of vec-
tor , again at the proper angle. (3) The vector sum is the vector that extends
from the tail of to the head of .
Properties. Vector addition, defined in this way, has two important proper-
ties. First, the order of addition does not matter. Adding to gives the sameb
:
a
:
b
:
a
:
s
:
a
:
b
:a
:
b
:
a
:
b
:
a
:
s
:
s
:a
:b
:,
a
:
Figure 3-1 (a) All three arrows have the
same magnitude and direction and thus
represent the same displacement. (b) All
three paths connecting the two points cor-
respond to the same displacement vector.
(a)
A'
B'
A"
B"
A
B
A
B
(b)
Figure 3-2 (a)AC is the vector sum of the
vectors AB and BC. (b) The same vectors
relabeled.
AC
B
(a)
A
ctual
path
Net displacement
is the vector sum
(b)
a
s
b
This is the
resulting vector,
from tail of a
to head of b.
To add a and b,
draw them
head to tail.
CHAPTER 3 VECTORS
42
result as adding to (Fig.3-3); that is,
(commutative law). (3-2)
Second, when there are more than two vectors, we can group them in any order
as we add them. Thus, if we want to add vectors ,, and , we can add and
first and then add their vector sum to . We can also add and first and then
add that sum to .We get the same result either way, as shown in Fig. 3-4. That is,
(associative law). (3-3)(a
:b
:)c
:a
:(b
:c
:)
a
:
c
:
b
:
c
:
b
:
a
:
c
:
b
:
a
:
a
:b
:b
:a
:
a
:
b
:
Figure 3-3 The two vectors and can be
added in either order; see Eq. 3-2.
b
:
a
:
a+b
b+a
FinishStart
Vector sum
a
a
b
b
You get the same vector
result for either order of
adding vectors.
Figure 3-4 The three vectors , , and can be grouped in any way as they are added; see
Eq. 3-3.
c
:
b
:
a
:
b+c
a+b
aa
cc
b
a+b
(a+b)+c
a+b+c
a+ (b+c)
b+c
You get the same vector result for
any order of adding the vectors.
Figure 3-5 The vectors and have theb
:
b
:
b
–b
Figure 3-6 (a) Vectors , , and .
(b) To subtract vector from vector ,
add vector to vector .a
:
b
:a
:
b
:b
:
b
:
a
:
d=a – b
(a)
(b)
Note head-to-tail
arrangement for
addition
a
a
b
–b
–b
Checkpoint 1
The magnitudes of displacements and are 3 m and 4 m, respectively, and .
Considering various orientations of and , what are (a) the maximum possible
magnitude for and (b) the minimum possible magnitude?
c
:
b
:
a
:
c
:a
:b
:
b
:
a
:
The vector is a vector with the same magnitude as but the opposite
direction (see Fig.3-5).Adding the two vectors in Fig.3-5 would yield
Thus, adding has the effect of subtracting . We use this property to define
the difference between two vectors: let . Then
(vector subtraction); (3-4)
that is, we find the difference vector by adding the vector to the vector .
Figure 3-6 shows how this is done geometrically.
As in the usual algebra, we can move a term that includes a vector symbol from
one side of a vector equation to the other,but we must change its sign. For example,
if we are given Eq. 3-4 and need to solve for , we can rearrange the equation as
Remember that, although we have used displacement vectors here, the rules
for addition and subtraction hold for vectors of all kinds, whether they represent
velocities, accelerations, or any other vector quantity. However, we can add
only vectors of the same kind. For example, we can add two displacements, or two
velocities, but adding a displacement and a velocity makes no sense. In the arith-
metic of scalars, that would be like trying to add 21 s and 12 m.
d
:b
:a
: or a
:d
:b
:.
a
:
a
:
b
:
d
:
d
:a
:b
:a
:(b
:)
d
:a
:b
:
b
:
b
:
b
:(b
:)0.
b
:
b
:
Components of Vectors
Adding vectors geometrically can be tedious. A neater and easier technique
involves algebra but requires that the vectors be placed on a rectangular coordi-
nate system.The xand yaxes are usually drawn in the plane of the page, as shown
same magnitude and opposite directions.
3-1 VECTORS AND THEIR COMPONENTS
in Fig.3-7a. The zaxis comes directly out of the page at the origin; we ignore it for
now and deal only with two-dimensional vectors.
Acomponent of a vector is the projection of the vector on an axis. In
Fig. 3-7a, for example, axis the component of vector on (or along) the xaxis and
ayis the component along the yaxis. To find the projection of a vector along an
axis, we draw perpendicular lines from the two ends of the vector to the axis, as
shown.The projection of a vector on an xaxis is its x component,and similarly the
projection on the yaxis is the y component. The process of finding the
components of a vector is called resolving the vector.
A component of a vector has the same direction (along an axis) as the vector.
In Fig. 3-7, axand ayare both positive because extends in the positive direction
of both axes. (Note the small arrowheads on the components, to indicate their di-
rection.) If we were to reverse vector , then both components would be negative
and their arrowheads would point toward negative xand y. Resolving vector in
Fig.3-8 yields a positive component bxand a negative component by.
In general, a vector has three components, although for the case of Fig. 3-7a
the component along the zaxis is zero.As Figs. 3-7aand bshow, if you shift a vec-
tor without changing its direction, its components do not change.
Finding the Components. We can find the components of in Fig. 3-7ageo-
metrically from the right triangle there:
axacos uand ayasin u, (3-5)
where uis the angle that the vector makes with the positive direction of the
xaxis, and ais the magnitude of . Figure 3-7cshows that and its xand ycom-
ponents form a right triangle. It also shows how we can reconstruct a vector from
its components: we arrange those components head to tail. Then we complete a
right triangle with the vector forming the hypotenuse, from the tail of one com-
ponent to the head of the other component.
Once a vector has been resolved into its components along a set of axes, the
components themselves can be used in place of the vector. For example, in
Fig. 3-7ais given (completely determined) by aand u. It can also be given by its
components axand ay. Both pairs of values contain the same information. If we
know a vector in component notation (axand ay) and want it in magnitude-angle
notation (aand u), we can use the equations
and tan (3-6)
to transform it.
In the more general three-dimensional case, we need a magnitude and two
angles (say, a,u, and f) or three components (ax,ay,and az) to specify a vector.
ay
ax
a2a2
xay
2
a
:
a
:
a
:
a
:
a
:
b
:
a
:
a
:
a
:
Figure 3-8 The component of on the
xaxis is positive, and that on the yaxis is
negative.
b
:
O
y (m)
θ
x(m)
bx= 7 m
by= –5 m
b
This is the x component
of the vector.
This is the y component
of the vector.
43
Figure 3-7 (a) The components axand ayof
vector . (b) The components are unchanged if
the vector is shifted, as long as the magnitude
and orientation are maintained. (c) The com-
ponents form the legs of a right triangle whose
hypotenuse is the magnitude of the vector.
a
:
y
x
Oax
ay
θ θ
(a) (b)
y
x
O
ax
ay
a a
θ
(c)
ay
ax
a
This is the y component
of the vector.
This is the x component
of the vector.
The components
and the vector
form a right triangle.
Checkpoint 2
In the figure, which of the indicated methods for combining the xand ycomponents of vector are proper to determine that vector?
a
:
y
x
ax
ay
(a)
a
y
x
ax
ay
(d)
a
y
x
ax
ay
(e)
a
x
ax
ay
y
(f)
a
y
x
ax
ay
(b)
a
y
x
ax
ay
(c)
a
CHAPTER 3 VECTORS
44
KEY IDEA
We are given the magnitude (215 km) and the angle (22° east
of due north) of a vector and need to find the components
of the vector.
Calculations: We draw an xy coordinate system with the
positive direction of xdue east and that of ydue north (Fig.
3-10). For convenience, the origin is placed at the airport.
(We don’t have to do this. We could shift and misalign the
coordinate system but, given a choice, why make the prob-
lem more difficult?) The airplane’s displacement points
from the origin to where the airplane is sighted.
To find the components of , we use Eq. 3-5 with u
68° (90° 22°):
dxdcos u(215 km)(cos 68°)
81 km (Answer)
dydsin u(215 km)(sin 68°)
199 km 2.0 102km. (Answer)
Thus, the airplane is 81 km east and 2.0 102km north of
the airport.
d
:
d
:
Sample Problem 3.02 Finding components, airplane flight
A small airplane leaves an airport on an overcast day and is
later sighted 215 km away, in a direction making an angle of
22° east of due north. This means that the direction is not
due north (directly toward the north) but is rotated 22° to-
ward the east from due north. How far east and north is the
airplane from the airport when sighted?
Additional examples, video, and practice available at WileyPLUS
Figure 3-10 A plane takes off from an airport at the origin and is
later sighted at P.
215 km
100
y
x
200
00 100
22°
θ
Distance (km)
Distance (km)
P
d
order, because their vector sum is the same for any order.
(Recall from Eq. 3-2 that vectors commute.) The order
shown in Fig. 3-9bis for the vector sum
Using the scale given in Fig.3-9a, we measure the length dof
this vector sum, finding
d4.8 m. (Answer)
d
:b
:a
:(c
:).
Sample Problem 3.01 Adding vectors in a drawing, orienteering
In an orienteering class, you have the goal of moving as far
(straight-line distance) from base camp as possible by
making three straight-line moves. You may use the follow-
ing displacements in any order: (a) , 2.0 km due east
(directly toward the east); (b) , 2.0 km 30° north of east
(at an angle of 30° toward the north from due east);
(c) , 1.0 km due west. Alternatively, you may substitute
either for or for . What is the greatest distance
you can be from base camp at the end of the third displace-
ment? (We are not concerned about the direction.)
Reasoning: Using a convenient scale, we draw vectors ,
, , , and as in Fig. 3-9a. We then mentally slide the
vectors over the page, connecting three of them at a time
in head-to-tail arrangements to find their vector sum .
The tail of the first vector represents base camp. The head
of the third vector represents the point at which you stop.
The vector sum extends from the tail of the first vector
to the head of the third vector. Its magnitude dis your dis-
tance from base camp. Our goal here is to maximize that
base-camp distance.
We find that distance dis greatest for a head-to-tail
arrangement of vectors , , and . They can be in any
c
:
b
:
a
:
d
:
d
:
c
:
b
:
c
:
b
:a
:
c
:
c
:
b
:
b
:
c
:
b
:a
:
Figure 3-9 (a) Displacement vectors; three are to be used. (b) Your
distance from base camp is greatest if you undergo
displacements , , and , in any order.c
:
b
:
a
:
30°
0 1
Scale of km
2
d = b+a – c
(
a
)(
b
)
a
a
c
bb
–b
–c
–c
This is the vector result
for adding those three
vectors in any order.
45
the xaxis. If it is measured relative to some other direc-
tion, then the trig functions in Eq. 3-5 may have to be in-
terchanged and the ratio in Eq. 3-6 may have to be
inverted. A safer method is to convert the angle to one
measured from the positive direction of the xaxis. In
WileyPLUS, the system expects you to report an angle of
direction like this (and positive if counterclockwise and
negative if clockwise).
Problem-Solving Tactics Angles, trig functions, and inverse trig functions
Tactic 1: Angles—Degrees and Radians Angles that are
measured relative to the positive direction of the xaxis are
positive if they are measured in the counterclockwise direc-
tion and negative if measured clockwise. For example, 210°
and 150° are the same angle.
Angles may be measured in degrees or radians (rad).To
relate the two measures, recall that a full circle is 360° and
2prad.To convert, say, 40° to radians, write
Tactic 2: Trig Functions You need to know the definitions
of the common trigonometric functions—sine, cosine, and
tangent—because they are part of the language of science
and engineering. They are given in Fig. 3-11 in a form that
does not depend on how the triangle is labeled.
You should also be able to sketch how the trig functions
vary with angle, as in Fig. 3-12, in order to be able to judge
whether a calculator result is reasonable. Even knowing
the signs of the functions in the various quadrants can be
of help.
Tactic 3: Inverse Trig Functions When the inverse trig
functions sin1, cos1, and tan1are taken on a calculator,
you must consider the reasonableness of the answer you
get, because there is usually another possible answer that
the calculator does not give. The range of operation for a
calculator in taking each inverse trig function is indicated
in Fig. 3-12. As an example, sin10.5 has associated angles
of 30° (which is displayed by the calculator, since 30° falls
within its range of operation) and 150°.To see both values,
draw a horizontal line through 0.5 in Fig. 3-12aand note
where it cuts the sine curve. How do you distinguish a cor-
rect answer? It is the one that seems more reasonable for
the given situation.
Tactic 4: Measuring Vector Angles The equations for
cos uand sin uin Eq. 3-5 and for tan uin Eq. 3-6 are valid
only if the angle is measured from the positive direction of
402
rad
3600.70 rad.
Figure 3-11 A triangle used to define the trigonometric
functions. See also Appendix E.
θ
Hypotenuse
Leg adjacent to
θ
Leg
opposite
θ
sin
θ
leg opposite
θ
hypotenuse
=
cos
θ
hypotenuse
=leg adjacent to
θ
tan
θ
=leg adjacent to
θ
leg opposite
θ
3-1 VECTORS AND THEIR COMPONENTS
Additional examples, video, and practice available at WileyPLUS
Figure 3-12 Three useful curves to remember. A calculator’s range
of operation for taking inverse trig functions is indicated by the
darker portions of the colored curves.
90° 270° –90°
+1
–1
IV I II III IV
Quadrants
(a)
0
sin
180° 360°
(b)
0
cos
90° 180° 270° 360° –90°
+1
–1
(c)
90° 270°–90°
+1
+2
–1
–2
tan
180° 360°0
CHAPTER 3 VECTORS
46
3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS
After reading this module, you should be able to . . .
3.06 Convert a vector between magnitude-angle and unit-
vector notations.
3.07 Add and subtract vectors in magnitude-angle notation
and in unit-vector notation.
3.08 Identify that, for a given vector, rotating the coordinate
system about the origin can change the vector’s compo-
nents but not the vector itself.
●Unit vectors , , and have magnitudes of unity and are
directed in the positive directions of the x,y, and zaxes,
respectively, in a right-handed coordinate system. We can
write a vector in terms of unit vectors as
axi
ˆayj
ˆazk
ˆ ,a
:
a
:
k
ˆ
j
ˆ
i
ˆin which , , and are the vector components of and
ax,ay, and azare its scalar components.
●To add vectors in component form, we use the rules
rxaxbxryaybyrzazbz.
Here and are the vectors to be added, and is the vector
sum. Note that we add components axis by axis.
r
:
b
:
a
:
a
:
azk
ˆ
ayj
ˆ
axi
ˆ
Learning Objectives
Key Ideas
ˆ
ˆ
y
x
Oaxi
ayj
θ
(a)
abxi
ˆ
ˆ
θ
Ox
y
byj
(b)
b
This is the x vector
component.
This is the y vector component.
Figure 3-14 (a) The vector components
of vector . (b) The vector components
of vector .
b
:
a
:
Unit Vectors
Aunit vector is a vector that has a magnitude of exactly 1 and points in a particu-
lar direction. It lacks both dimension and unit. Its sole purpose is to point—that
is, to specify a direction.The unit vectors in the positive directions of the x,y, and
zaxes are labeled , , and , where the hat is used instead of an overhead arrow
as for other vectors (Fig. 3-13).The arrangement of axes in Fig. 3-13 is said to be a
right-handed coordinate system. The system remains right-handed if it is rotated
rigidly.We use such coordinate systems exclusively in this book.
Unit vectors are very useful for expressing other vectors; for example, we can
express and of Figs. 3-7 and 3-8 as
(3-7)
and . (3-8)
These two equations are illustrated in Fig. 3-14.The quantities axand ayare vec-
tors, called the vector components of .The quantities axand ayare scalars, called
the scalar components of (or, as before,simply its components).a
:
a
:
j
ˆ
i
ˆ
b
:bxi
ˆbyj
ˆ
a
:axi
ˆayj
ˆ
b
:
a
:
ˆk
ˆ
j
ˆ
i
ˆ
Adding Vectors by Components
We can add vectors geometrically on a sketch or directly on a vector-capable
calculator.A third way is to combine their components axis by axis.
Figure 3.13 Unit vectors i
ˆ, , and define the
directions of a right-handed coordinate
system.
k
ˆ
j
ˆ
y
x
z
j
ˆ
i
ˆ
k
ˆ
The unit vectors point
along axes.
47
To start, consider the statement
,(3-9)
which says that the vector is the same as the vector . Thus, each
component of must be the same as the corresponding component of :
rxaxbx(3-10)
ryayby(3-11)
rzazbz. (3-12)
In other words, two vectors must be equal if their corresponding components are
equal. Equations 3-9 to 3-12 tell us that to add vectors and , we must (1) re-
solve the vectors into their scalar components; (2) combine these scalar compo-
nents, axis by axis, to get the components of the sum ; and (3) combine
the components of to get itself. We have a choice in step 3. We can express
in unit-vector notation or in magnitude-angle notation.
This procedure for adding vectors by components also applies to vector
subtractions. Recall that a subtraction such as can be rewritten as an
addition .To subtract, we add and by components, to get
dxaxbx,dyayby, and dzazbz,
where . (3-13)d
:dxi
ˆdyj
ˆdzk
ˆ
b
:
a
:
d
:a
:(b
:)
d
:a
:b
:
r
:
r
:
r
:
r
:
b
:
a
:
(a
:b
:)r
:
(a
:b
:)r
:
r
:a
:b
:
Checkpoint 3
(a) In the figure here, what are the signs of the x
components of and ? (b) What are the signs of
the ycomponents of and ? (c) What are thed2
:
d1
:
d2
:
d1
:
y
x
d2
d1
3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS
Vectors and the Laws of Physics
So far, in every figure that includes a coordinate system, the xand yaxes are par-
allel to the edges of the book page. Thus, when a vector is included, its compo-
nents axand ayare also parallel to the edges (as in Fig. 3-15a).The only reason for
that orientation of the axes is that it looks “proper”; there is no deeper reason.
We could, instead, rotate the axes (but not the vector ) through an angle fas ina
:
a
:
Figure 3-15 (a) The vector and its
components. (b) The same vector, with the
axes of the coordinate system rotated
through an angle f.
a
:
a
y
x
ax
ay
θ
(a)
O
a
y
x
a'
x
x'
(b)
θ
a'
y
φ
O
y'
'
Rotating the axes
changes the components
but not the vector.
Fig. 3-15b, in which case the components would have new values, call them a
xand
a
y. Since there are an infinite number of choices of f, there are an infinite num-
ber of different pairs of components for .
Which then is the “right” pair of components? The answer is that they are all
equally valid because each pair (with its axes) just gives us a different way of de-
scribing the same vector ; all produce the same magnitude and direction for the
vector.In Fig. 3-15 we have
(3-14)
and
uuf. (3-15)
The point is that we have great freedom in choosing a coordinate system, be-
cause the relations among vectors do not depend on the location of the origin or
on the orientation of the axes.This is also true of the relations of physics; they are
all independent of the choice of coordinate system.Add to that the simplicity and
richness of the language of vectors and you can see why the laws of physics are
almost always presented in that language: one equation, like Eq. 3-9, can repre-
sent three (or even more) relations, like Eqs. 3-10, 3-11, and 3-12.
a2a2
xa2
y2a2
xa2
y
a
:
a
:
signs of the xand ycomponents of ?d2
:
d1
:
CHAPTER 3 VECTORS
48
Calculations: To evaluate Eqs. 3-16 and 3-17, we find the xand
ycomponents of each displacement. As an example, the com-
ponents for the first displacement are shown in Fig. 3-16c.We
draw similar diagrams for the other two displacements and
then we apply the xpart of Eq. 3-5 to each displacement, using
angles relative to the positive direction of the xaxis:
dlx(6.00 m) cos 40° 4.60 m
d2x(8.00 m) cos (60°) 4.00 m
d3x(5.00 m) cos 0° 5.00 m.
Equation 3-16 then gives us
dnet,x4.60 m 4.00 m 5.00 m
13.60 m.
Similarly, to evaluate Eq. 3-17, we apply the ypart of Eq. 3-5
to each displacement:
dly(6.00 m) sin 40° = 3.86 m
d2y(8.00 m) sin (60°) = 6.93 m
d3y(5.00 m) sin 0° 0 m.
Equation 3-17 then gives us
dnet,y3.86 m 6.93 m 0 m
3.07 m.
Next we use these components of net to construct the vec-
tor as shown in Fig. 3-16d: the components are in a head-to-
tail arrangement and form the legs of a right triangle, and
d
:
Sample Problem 3.03 Searching through a hedge maze
A hedge maze is a maze formed by tall rows of hedge.
After entering, you search for the center point and then
for the exit. Figure 3-16ashows the entrance to such a
maze and the first two choices we make at the junctions
we encounter in moving from point ito point c. We un-
dergo three displacements as indicated in the overhead
view of Fig. 3-16b:
d16.00 m
140°
d28.00 m
230°
d35.00 m
30°,
where the last segment is parallel to the superimposed
xaxis. When we reach point c, what are the magnitude and
angle of our net displacement net from point i?
KEY IDEAS
(1) To find the net displacement net, we need to sum the
three individual displacement vectors:
net 123.
(2) To do this, we first evaluate this sum for the xcompo-
nents alone,
dnet,xdlxd2xd3x, (3-16)
and then the ycomponents alone,
dnet,yd1yd2yd3y. (3-17)
(3) Finally, we construct net from its xand ycomponents.d
:
d
:
d
:
d
:
d
:
d
:
d
:
Figure 3-16 (a) Three displacements through a hedge maze. (b) The displacement vectors. (c) The first displacement vector and its
components. (d) The net displacement vector and its components.
(a)
y
x
d1y
d1x
(c)
a
bc
i(b)
y
x
a
bc
iu1
u2
y
x
dnet,x
dnet,y
c
(d)
d1d2
d3
d1
i
dnet
Three
vectors First
vector
Net
vector
49
the vector forms the hypotenuse.We find the magnitude and
angle of net with Eq. 3-6. The magnitude is
dnet (3-18)
13.9 m. (Answer)
To find the angle (measured from the positive direction of x),
we take an inverse tangent:
tan1(3-19)
tan112.7°. (Answer)
The angle is negative because it is measured clockwise from
positive x.We must always be alert when we take an inverse
–3.07 m
13.60 m
dnet,y
dnet,x
2(13.60 m)2(3.07 m)2
2d2
net,xd2
net,y
d
:tangent on a calculator. The answer it displays is mathe-
matically correct but it may not be the correct answer for
the physical situation. In those cases, we have to add 180°
to the displayed answer, to reverse the vector. To check,
we always need to draw the vector and its components as
we did in Fig. 3-16d. In our physical situation, the figure
shows us that
12.7° is a reasonable answer, whereas
12.7° 180° 167° is clearly not.
We can see all this on the graph of tangent versus angle
in Fig. 3-12c. In our maze problem, the argument of the in-
verse tangent is 3.07/13.60, or 0.226. On the graph draw
a horizontal line through that value on the vertical axis. The
line cuts through the darker plotted branch at 12.7° and
also through the lighter branch at 167°. The first cut is what
a calculator displays.
3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS
KEY IDEA
We can add the three vectors by components, axis by axis,
and then combine the components to write the vector
sum .
Calculations: For the xaxis, we add the xcomponents of
and to get the xcomponent of the vector sum :
rxaxbxcx
4.2 m 1.6 m 02.6 m.
Similarly, for the yaxis,
ryaybycy
1.5 m 2.9 m 3.7 m 2.3 m.
We then combine these components of to write the vector
in unit-vector notation:
(Answer)
where (2.6 m)i
ˆis the vector component of along the xaxis
and (2.3 m)j
ˆis that along the yaxis. Figure 3-17bshows
one way to arrange these vector components to form .
(Can you sketch the other way?)
We can also answer the question by giving the magnitude
and an angle for .From Eq.3-6,the magnitude is
(Answer)
and the angle (measured from the xdirection) is
(Answer)
where the minus sign means clockwise.
tan1
2.3 m
2.6 m
41,
r2(2.6 m)2(2.3 m)23.5 m
r
:
r
:
r
:
r
:(2.6 m)i
ˆ(2.3 m)j
ˆ,
r
:
r
:
c
:,b
:,
a
:,
r
:
Sample Problem 3.04 Adding vectors, unit-vector components
Figure 3-17ashows the following three vectors:
and
What is their vector sum which is also shown?r
:
c
:(3.7 m)j
ˆ.
b
:(1.6 m)i
ˆ(2.9 m)j
ˆ,
a
:(4.2 m)i
ˆ(1.5 m)j
ˆ,
Additional examples, video, and practice available at WileyPLUS
x
y
–1 3 4 –2–3 2
–3
–2
–1
1
x
y
–1 3 4 –2–3 2
–3
–2
–1
2
3
1
1
(a)
2.6i
(b)
r
r
a
c
b
ˆ
–2.3j
ˆ
To add these vectors,
find their net x component
and their net y component.
Then arrange the net
components head to tail.
This is the result of the addition.
Figure 3-17 Vector is the vector sum of the other three vectors.r
:
CHAPTER 3 VECTORS
50
Multiplying Vectors*
There are three ways in which vectors can be multiplied, but none is exactly like
the usual algebraic multiplication. As you read this material, keep in mind that a
vector-capable calculator will help you multiply vectors only if you understand
the basic rules of that multiplication.
Multiplying a Vector by a Scalar
If we multiply a vector by a scalar s, we get a new vector. Its magnitude is
the product of the magnitude of and the absolute value of s. Its direction is the
direction of if sis positive but the opposite direction if sis negative.To divide
by s, we multiply by 1/s.
Multiplying a Vector by a Vector
There are two ways to multiply a vector by a vector: one way produces a scalar
(called the scalar product), and the other produces a new vector (called the vector
product). (Students commonly confuse the two ways.)
a
:
a
:
a
:
a
:
a
:
Key Ideas
●The vector (or cross) product of two vectors and is
written and is a vector whose magnitude cis given by
cab sin
,
in which
is the smaller of the angles between the directions
of and . The direction of is perpendicular to the plane
defined by and and is given by a right-hand rule, as shown
in Fig. 3-19. Note that (). In unit-vector
notation,
which we may expand with the distributive law.
●In nested products, where one product is buried inside an-
other, follow the normal algebraic procedure by starting with
the innermost product and working outward.
(bxi
ˆbyj
ˆbzk
ˆ ),(axi
ˆayj
ˆazk
ˆ )b
:
a
:
a
:
b
:
b
:
a
:
b
:
a
:
c
:
b
:
a
:
c
:
b
:
a
:
b
:
a
:
*This material will not be employed until later (Chapter 7 for scalar products and Chapter 11 for vec-
tor products), and so your instructor may wish to postpone it.
●The product of a scalar sand a vector is a new vector
whose magnitude is and whose direction is the same as
that of if sis positive, and opposite that of if sis negative.
To divide by s, multiply by 1/s.
●The scalar (or dot) product of two vectors and is writ-
ten and is the scalar quantity given by
ab cos
,
in which
is the angle between the directions of and .
A scalar product is the product of the magnitude of one vec-
tor and the scalar component of the second vector along the
direction of the first vector. In unit-vector notation,
which may be expanded according to the distributive law.
Note that .a
:
b
:
b
:
a
:
(bxi
ˆbyj
ˆbzk
ˆ ),(axi
ˆayj
ˆazk
ˆ )b
:
a
:
b
:
a
:
b
:
a
:
b
:
a
:
b
:
a
:
v
:
v
:
v
:
v
:
sv
v
:
3-3 MULTIPLYING VECTORS
Learning Objectives
3.13 Given two vectors, use a dot product to find how much
of one vector lies along the other vector.
3.14 Find the cross product of two vectors in magnitude-
angle and unit-vector notations.
3.15 Use the right-hand rule to find the direction of the vector
that results from a cross product.
3.16 In nested products, where one product is buried inside
another, follow the normal algebraic procedure by starting
with the innermost product and working outward.
After reading this module, you should be able to . . .
3.09 Multiply vectors by scalars.
3.10 Identify that multiplying a vector by a scalar gives a vec-
tor, taking the dot (or scalar) product of two vectors gives a
scalar, and taking the cross (or vector) product gives a new
vector that is perpendicular to the original two.
3.11 Find the dot product of two vectors in magnitude-angle
notation and in unit-vector notation.
3.12 Find the angle between two vectors by taking their dot prod-
uct in both magnitude-angle notation and unit-vector notation.
3-3 M U LT I P LY I N G V E C TO R S
If the angle between two vectors is 0°, the component of one vector along the
other is maximum, and so also is the dot product of the vectors. If, instead, is 90°,
the component of one vector along the other is zero, and so is the dot product.
The Scalar Product
The scalar product of the vectors and in Fig. 3-18ais written as and
defined to be
ab cos f, (3-20)
where ais the magnitude of , bis the magnitude of , and is the angle between
and (or,more properly, between the directions of and ).There are actually
two such angles: and 360° . Either can be used in Eq. 3-20, because their
cosines are the same.
Note that there are only scalars on the right side of Eq. 3-20 (including the
value of cos ).Thus on the left side represents a scalar quantity. Because of
the notation, is also known as the dot product and is spoken as “a dot b.”
A dot product can be regarded as the product of two quantities: (1) the mag-
nitude of one of the vectors and (2) the scalar component of the second vector
along the direction of the first vector. For example, in Fig. 3-18b, has a scalar
component acos along the direction of ; note that a perpendicular dropped
from the head of onto determines that component. Similarly, has a scalar
component bcos along the direction of .a
:
b
:
b
:
a
:
b
:
a
:
b
:
a
:
b
:
a
:
b
:
a
:
b
:
a
:
b
:
a
:
a
:b
:
a
:b
:
b
:
a
:
51
Figure 3-18 (a) Two vectors
and , with an angle fbetween
them. (b) Each vector has a
component along the direction
of the other vector.
b
:a
:
a
a
b
b
φ
(a)
(b)
Component of b
along direction of
a is b cos
φ
Component of a
along direction of
b is a cos
φ
φ
Multiplying these gives
the dot product.
Or multiplying these
gives the dot product.
Equation 3-20 can be rewritten as follows to emphasize the components:
(acos f)(b)(a)(bcos f). (3-21)
The commutative law applies to a scalar product, so we can write
.
When two vectors are in unit-vector notation,we write their dot product as
(axayaz)(bxbybz), (3-22)
which we can expand according to the distributive law: Each vector component
of the first vector is to be dotted with each vector component of the second vec-
tor.By doing so, we can show that
axbxaybyazbz. (3-23)b
:
a
:
k
ˆ
j
ˆ
i
ˆ
k
ˆ
j
ˆ
i
ˆ
b
:
a
:
a
:
b
:
b
:
a
:
b
:
a
:
CHAPTER 3 VECTORS
52
If and are parallel or antiparallel, 0. The magnitude of , which can
be written as , is maximum when and are perpendicular to each other.
b
:
a
:
a
:b
:
b
:
a
:
b
:
a
:
b
:
a
:
where fis the smaller of the two angles between and . (You must use theb
:
a
:
The direction of is perpendicular to the plane that contains and .b
:
a
:
c
:
Checkpoint 4
Vectors and have magnitudes of 3 units and 4 units, respectively.What is the
angle between the directions of and if equals (a) zero, (b) 12 units, and
(c) 12 units?
D
:
C
:D
:
C
:
D
:
C
:
The Vector Product
The vector product of and , written , produces a third vector whose
magnitude is
cab sin f, (3-24)
c
:
b
:
a
:
b
:
a
:
them is 90°.) Also, we used the right-hand rule to get the direction of as
being in the positive direction of the zaxis (thus in the direction of ).k
ˆj
ˆ
i
ˆ
smaller of the two angles between the vectors because sin fand sin(360° f)
differ in algebraic sign.) Because of the notation, is also known as the cross
product, and in speech it is “a cross b.”
b
:
a
:
Figure 3-19ashows how to determine the direction of with what is
known as a right-hand rule. Place the vectors and tail to tail without altering
their orientations, and imagine a line that is perpendicular to their plane where
they meet. Pretend to place your right hand around that line in such a way that
your fingers would sweep into through the smaller angle between them.Your
outstretched thumb points in the direction of .
The order of the vector multiplication is important. In Fig. 3-19b, we are
determining the direction of , so the fingers are placed to sweep
into through the smaller angle. The thumb ends up in the opposite direction
from previously, and so it must be that ; that is,
. (3-25)
In other words, the commutative law does not apply to a vector product.
In unit-vector notation, we write
(axayaz)(bxbybz), (3-26)
which can be expanded according to the distributive law; that is, each component
of the first vector is to be crossed with each component of the second vector.The
cross products of unit vectors are given in Appendix E (see “Products of
Vectors”). For example, in the expansion of Eq. 3-26, we have
axbxaxbx()0,
because the two unit vectors and are parallel and thus have a zero cross prod-
uct. Similarly, we have
axbyaxby()axby.
In the last step we used Eq. 3-24 to evaluate the magnitude of as unity.
(These vectors and each have a magnitude of unity, and the angle betweenj
ˆ
i
ˆj
ˆ
i
ˆ
k
ˆ
j
ˆ
i
ˆ
j
ˆ
i
ˆ
i
ˆ
i
ˆ
i
ˆ
i
ˆ
i
ˆ
i
ˆ
k
ˆ
j
ˆ
i
ˆ
k
ˆ
j
ˆ
i
ˆ
b
:
a
:
b
:a
:(a
:b
:)
c
:c
:
a
:
b
:
c
:b
:a
:
c
:
b
:
a
:
b
:
a
:
b
:
a
:
c
:
53
3-3 M U LT I P LY I N G V E C TO R S
Continuing to expand Eq. 3-26, you can show that
(aybzbyaz)(azbxbzax)(axbybxay) . (3-27)
A determinant (Appendix E) or a vector-capable calculator can also be used.
To check whether any xyz coordinate system is a right-handed coordinate
system, use the right-hand rule for the cross product with that system. If
your fingers sweep (positive direction of x) into (positive direction of y) with
the outstretched thumb pointing in the positive direction of z(not the negative
direction), then the system is right-handed.
j
ˆ
i
ˆk
ˆ
j
ˆ
i
ˆ
k
ˆ
j
ˆ
i
ˆ
b
:
a
:
Checkpoint 5
Vectors and have magnitudes of 3 units and 4 units, respectively.What is the an-
gle between the directions of and if the magnitude of the vector product
is (a) zero and (b) 12 units?
D
:
C
:D
:
C
:
D
:
C
:
Figure 3-19 Illustration of the right-hand rule for vector products. (a) Sweep vector into vector with the fingers of your right hand.
Your outstretched thumb shows the direction of vector . (b) Showing that is the reverse of .a
:b
:
b
:a
:
c
:a
:b
:b
:
a
:
a
bbb
c
a
b
aa
(a)
(b)
c
A
CHAPTER 3 VECTORS
54
gives the direction of .Thus, as shown in the figure, lies in
the xy plane. Because its direction is perpendicular to the
direction of (a cross product always gives a perpendicular
vector), it is at an angle of
250° 90° 160° (Answer)
from the positive direction of the xaxis.
a
:
c
:
c
:
Sample Problem 3.06 Cross product, right-hand rule
In Fig. 3-20, vector lies in the xy plane, has a magnitude of
18 units, and points in a direction 250° from the positive di-
rection of the xaxis. Also, vector has a magnitude of
12 units and points in the positive direction of the zaxis.What
is the vector product ?
KEY IDEA
When we have two vectors in magnitude-angle notation, we
find the magnitude of their cross product with Eq. 3-24 and
the direction of their cross product with the right-hand rule
of Fig.3-19.
Calculations: For the magnitude we write
cab sin f(18)(12)(sin 90°) 216. (Answer)
To determine the direction in Fig. 3-20, imagine placing the
fingers of your right hand around a line perpendicular to the
plane of and (the line on which is shown) such that
your fingers sweep into . Your outstretched thumb thenb
:
a
:
c
:
b
:
a
:
b
:
a
:
c
:
b
:
a
:
Figure 3-20 Vector (in the xy plane) is the vector (or cross)
product of vectors and .b
:
a
:
c
:
z
250°
160°
y
x
ab
c = ab
This is the resulting
vector, perpendicular to
both a and b.
Sweep a into b.
Calculations: Here we write
(3 4)(23)
3(2)33(4)(2)
(4)3.k
ˆ
j
ˆ
i
ˆ
j
ˆ
k
ˆ
i
ˆ
i
ˆ
i
ˆ
k
ˆ
i
ˆ
j
ˆ
i
ˆ
c
:
Sample Problem 3.07 Cross product, unit-vector notation
If 34 and 23 ,what is ?
KEY IDEA
When two vectors are in unit-vector notation, we can find
their cross product by using the distributive law.
b
:
a
:
c
:
k
ˆ
i
ˆ
b
:
j
ˆ
i
ˆ
a
:
We can separately evaluate the left side of Eq. 3-28 by
writing the vectors in unit-vector notation and using the
distributive law:
(3.0 4.0 )(2.0 3.0 )
(3.0 )(2.0 ) (3.0 )(3.0 )
(4.0 )(2.0 ) (4.0 )(3.0 ).
We next apply Eq. 3-20 to each term in this last expression.
The angle between the unit vectors in the first term ( and ) is
0°,and in the other terms it is 90°.We then have
(6.0)(1) (9.0)(0) (8.0)(0) (12)(0)
6.0.
Substituting this result and the results of Eqs. 3-29 and 3-30
into Eq. 3-28 yields
6.0 (5.00)(3.61) cos f,
so (Answer)
cos16.0
(5.00)(3.61) 109110.
b
:
a
:
i
ˆ
i
ˆ
k
ˆ
j
ˆ
i
ˆ
j
ˆ
k
ˆ
i
ˆ
i
ˆ
i
ˆ
k
ˆ
i
ˆ
j
ˆ
i
ˆ
b
:
a
:
Sample Problem 3.05 Angle between two vectors using dot products
What is the angle between 3.0 4.0 and
2.0 3.0 ? (Caution: Although many of the following
steps can be bypassed with a vector-capable calculator, you
will learn more about scalar products if, at least here, you
use these steps.)
KEY IDEA
The angle between the directions of two vectors is included
in the definition of their scalar product (Eq. 3-20):
ab cos f. (3-28)
Calculations: In Eq. 3-28, ais the magnitude of , or
(3-29)
and bis the magnitude of , or
(3-30)b2(2.0)2 3.02 3.61.
b
:
a23.02(4.0)25.00,
a
:
b
:
a
:
k
ˆ
i
ˆ
b
:j
ˆ
i
ˆa
:
55
REVIEW & SUMMARY
Scalars and Vectors Scalars, such as temperature, have magni-
tude only. They are specified by a number with a unit (10°C) and
obey the rules of arithmetic and ordinary algebra. Vectors, such as
displacement, have both magnitude and direction (5 m, north) and
obey the rules of vector algebra.
Adding Vectors Geometrically Two vectors and may
be added geometrically by drawing them to a common scale
and placing them head to tail. The vector connecting the tail of
the first to the head of the second is the vector sum . To
subtract from , reverse the direction of to get ; then
add to .Vector addition is commutative
and obeys the associative law
.
Components of a Vector The (scalar) components axand ayof
any two-dimensional vector along the coordinate axes are found
by dropping perpendicular lines from the ends of onto the coor-
dinate axes.The components are given by
axacos uand ayasin u, (3-5)
where uis the angle between the positive direction of the xaxis
and the direction of . The algebraic sign of a component indi-
cates its direction along the associated axis. Given its compo-
nents, we can find the magnitude and orientation (direction) of
the vector by using
and
Unit-Vector Notation Unit vectors , , and have magnitudes of
unity and are directed in the positive directions of the x, y, and z
axes, respectively, in a right-handed coordinate system (as defined
by the vector products of the unit vectors).We can write a vector
in terms of unit vectors as
axayaz, (3-7)
in which ax,ay, and azare the vector components of and ax,ay,
and azare its scalar components.
a
:
k
ˆ
j
ˆ
i
ˆ
k
ˆ
j
ˆ
i
ˆ
a
:
a
:
k
ˆ
j
ˆ
i
ˆ
a2a2
xa2
y
a
:
a
:
a
:
a
:
(a
:b
:)c
:a
:(b
:c
:)
a
:b
:b
:a
:
a
:
b
:b
:
b
:
a
:
b
:s
:
b
:
a
:
Review & Summary
Adding Vectors in Component Form To add vectors in com-
ponent form, we use the rules
rxaxbxryaybyrzazbz. (3-10 to 3-12)
Here and are the vectors to be added, and is the vector sum.
Note that we add components axis by axis.We can then express the
sum in unit-vector notation or magnitude-angle notation.
Product of a Scalar and a Vector The product of a scalar sand
a vector is a new vector whose magnitude is sv and whose direc-
tion is the same as that of if sis positive, and opposite that of if
sis negative. (The negative sign reverses the vector.) To divide by
s, multiply by 1/s.
The Scalar Product The scalar (or dot)product of two vectors
and is written and is the scalar quantity given by
ab cos f, (3-20)
in which fis the angle between the directions of and . A scalar
product is the product of the magnitude of one vector and the
scalar component of the second vector along the direction of the
first vector. Note that which means that the scalar
product obeys the commutative law.
In unit-vector notation,
(axayaz)(bxbybz), (3-22)
which may be expanded according to the distributive law.
The Vector Product The vector (or cross)product of two vectors
and is written and is a vector whose magnitude cis
given by
cab sin f, (3-24)
in which fis the smaller of the angles between the directions of
and . The direction of is perpendicular to the plane
defined by and and is given by a right-hand rule, as shown in
Fig. 3-19. Note that (), which means that the vec-
tor product does not obey the commutative law.
In unit-vector notation,
(axayaz)(bxbybz), (3-26)
which we may expand with the distributive law.
k
ˆ
j
ˆ
i
ˆ
k
ˆ
j
ˆ
i
ˆ
b
:
a
:
a
:
b
:
b
:
a
:
b
:
a
:
c
:
b
:a
:
c
:
b
:
a
:
b
:
a
:
k
ˆ
j
ˆ
i
ˆ
k
ˆ
j
ˆ
i
ˆ
b
:
a
:
a
:,b
:
b
:
a
:
b
:
a
:
b
:
a
:
b
:
a
:
b
:a
:
v
:
v
:
v
:
v
:
v
:
r
:
b
:
a
:
Additional examples, video, and practice available at WileyPLUS
We next evaluate each term with Eq. 3-24, finding the
direction with the right-hand rule. For the first term here,
the angle fbetween the two vectors being crossed is 0. For
the other terms, fis 90°.We find
6(0) 9()8()12
12 98 . (Answer)k
ˆ
j
ˆ
i
ˆ
i
ˆ
k
ˆ
j
ˆ
c
:
This vector is perpendicular to both and , a fact youb
:
a
:
c
:
can check by showing that = 0 and = 0; that is, there
is no component of along the direction of either or .
In general: A cross product gives a perpendicular
vector, two perpendicular vectors have a zero dot prod-
uct, and two vectors along the same axis have a zero
cross product.
b
:
a
:
c
:
b
:
c
:
a
:
c
:
(3-2)
(3-3)
tan
ay
ax
(3-6)
CHAPTER 3 VECTORS
56
10 Figure 3-25 shows vector and
four other vectors that have the same
magnitude but differ in orientation.
(a) Which of those other four vectors
have the same dot product with ? (b)
Which have a negative dot product
with ?
11 In a game held within a three-
dimensional maze, you must move
your game piece from start, at xyz co-
ordinates (0, 0, 0), to finish, at coordinates (2 cm, 4 cm, 4 cm).
The game piece can undergo only the displacements (in centime-
ters) given below. If, along the way, the game piece lands at coordi-
nates (5 cm, 1 cm, 1 cm) or (5 cm, 2 cm, 1 cm), you lose the
game. Which displacements and in what sequence will get your
game piece to finish?
723232
24353.
12 The xand ycomponents of four vectors , , , and are given
below. For which vectors will your calculator give you the correct an-
gle uwhen you use it to find uwith Eq. 3-6? Answer first by examin-
ing Fig.3-12,and then check your answers with your calculator.
ax3ay 3cx3cy3
bx3by3dx3dy3.
13 Which of the following are correct (meaningful) vector
expressions? What is wrong with any incorrect expression?
(a) () (f) ()
(b) () (g) 5
(c) () (h) 5 ()
(d) () (i) 5 ()
(e) () (j) ( )()C
:
B
:
B
:
A
:
C
:
B
:
A
:
C
:
B
:
C
:
B
:
A
:
C
:
B
:
C
:
B
:
A
:
A
:
C
:
B
:
A
:
C
:
B
:
A
:
C
:
B
:
A
:
d
:
c
:
b
:
a
:
k
ˆ
j
ˆ
i
ˆ
s
:
k
ˆ
j
ˆ
i
ˆ
q
:
k
ˆ
j
ˆ
i
ˆ
r
:
k
ˆ
j
ˆ
i
ˆ
p
:
A
:
A
:
A
:
B
A
C
E
D
θ
θ θ
θ
Figure 3-25 Question 10.
FF
F
v
v v xxx
zzz
yyy
(1) (2) (3)
Figure 3-24 Question 9.
Figure 3-23 Question 5.
Questions
1Can the sum of the magnitudes
of two vectors ever be equal to the
magnitude of the sum of the same
two vectors? If no, why not? If yes,
when?
2The two vectors shown in Fig. 3-21
lie in an xy plane. What are the signs
of the xand ycomponents, respec-
tively, of (a) , (b) , and
(c) ?
3Being part of the “Gators,” the
University of Florida golfing team
must play on a putting green with an
alligator pit. Figure 3-22 shows an
overhead view of one putting chal-
lenge of the team; an xy coordinate
system is superimposed. Team mem-
bers must putt from the origin to the
hole, which is at xy coordinates (8 m,
12 m), but they can putt the golf ball
using only one or more of the fol-
lowing displacements, one or more
times:
,.
The pit is at coordinates (8 m, 6 m). If a team member putts the
ball into or through the pit, the member is automatically trans-
ferred to Florida State University, the arch rival. What sequence
of displacements should a team member use to avoid the pit and
the school transfer?
4Equation 3-2 shows that the addition of two vectors and is
commutative. Does that mean subtraction is commutative, so that
?
5Which of the arrangements of axes in Fig. 3-23 can be labeled
“right-handed coordinate system”? As usual, each axis label indi-
cates the positive side of the axis.
a
:
b
:
b
:
a
:
b
:
a
:
d3
:(8 m)i
ˆ
d2
:(6 m)ˆ
j,d1
:(8 m)i
ˆ(6 m)j
ˆ
d2
:d1
:d1
:d2
:
d1
:d2
:
6Describe two vectors and such that
(a) and abc;
(b) ;
(c) and a2b2c2.
7If (), does (a) () (), (b)
() , and (c) () ?
8If , must equal ?
9If q( ) and is perpendicular to , then what is the
direction of in the three situations shown in Fig. 3-24 when con-
stant qis (a) positive and (b) negative?
B
:B
:
v
:
B
:
v
:
F
:
c
:
b
:
c
:
a
:
b
:
a
:
b
:
a
:
d
:
c
:
c
:
d
:
b
:a
:
b
:
c
:
d
:
a
:
c
:
b
:
a
:
d
:
c
:
b
:
a
:
b
:
a
:
b
:
a
:
c
:
b
:
a
:
b
:
a
:
y
x
d2
d1
Figure 3-21 Question 2.
Hole
Gator
pit
y
x
Figure 3-22 Question 3.
z
y
x
(a )
y
z
x
(b )
x
z
y
(c )
yy
y
x
z
(d )
x
z
(e )
z
x
( f )
tors and in Fig. 3-28 have equal
magnitudes of 10.0 m and the angles
are 30° and 105°. Find the
(a) xand (b) ycomponents of their
vector sum , (c) the magnitude of ,
and (d) the angle makes with the
positive direction of the xaxis.
•16 For the displacement vectors
and
, give in
(a) unit-vector notation, and as (b) a
magnitude and (c) an angle (rela-
tive to ). Now give in (d) unit-vector notation, and as (e) a
magnitude and (f) an angle.
•17 Three vectors , , and each have a magnitude of
50 m and lie in an xy plane.Their directions relative to the positive
direction of the xaxis are 30°, 195°, and 315°, respectively.What are
(a) the magnitude and (b) the angle of the vector , and
(c) the magnitude and (d) the angle of ? What are the
(e) magnitude and (f) angle of a fourth vector such that
?
•18 In the sum , vector has a magnitude of 12.0 m
and is angled 40.0° counterclockwise from the direction, and vec-
tor has a magnitude of 15.0 m and is angled 20.0° counterclock-
wise from the direction.What are (a) the magnitude and (b) the
angle (relative to ) of ?
•19 In a game of lawn chess, where pieces are moved between
the centers of squares that are each 1.00 m on edge, a knight is
moved in the following way: (1) two squares forward, one square
rightward; (2) two squares leftward, one square forward; (3) two
squares forward, one square leftward. What are (a) the magnitude
and (b) the angle (relative to “forward”) of the knight’s overall dis-
placement for the series of three moves?
B
:
x
x
C
:x
A
:
A
:B
:C
:
(a
:b
:)(c
:d
:)0
d
:
a
:b
:c
:
a
:b
:c
:
c
:
b
:
a
:
ILW
b
:a
:
i
ˆ
a
:b
:
(5.0 m)i
ˆ(2.0 m)j
ˆb
:a
:(3.0 m)i
ˆ(4.0 m)j
ˆ
r
:
r
:
r
:
2
1
b
:
a
:
Module 3-1 Vectors and Their Components
•1 What are (a) the xcomponent and (b) the ycomponent of a
vector in the xy plane if its direction is 250°
counterclockwise from the positive direction
of the xaxis and its magnitude is 7.3 m?
•2 A displacement vector in the xy plane
is 15 m long and directed at angle u30° in
Fig. 3-26. Determine (a) the xcomponent
and (b) the ycomponent of the vector.
•3 The xcomponent of vector is
25.0 m and the ycomponent is 40.0 m. (a) What is the magni-
tude of ? (b) What is the angle between the direction of and
the positive direction of x?
•4 Express the following angles in radians: (a) 20.0°, (b) 50.0°,
(c) 100°. Convert the following angles to degrees: (d) 0.330 rad,
(e) 2.10 rad, (f) 7.70 rad.
•5 A ship sets out to sail to a point 120 km due north. An unex-
pected storm blows the ship to a point 100 km due east of its
starting point. (a) How far and (b) in what direction must it now
sail to reach its original destination?
•6 In Fig. 3-27, a heavy piece of
machinery is raised by sliding it a
distance d12.5 m along a plank
oriented at angle u20.0° to the
horizontal. How far is it moved
(a) vertically and (b) horizontally?
•7 Consider two displacements,
one of magnitude 3 m and another
of magnitude 4 m. Show how the
displacement vectors may be combined to get a resultant displace-
ment of magnitude (a) 7 m, (b) 1 m, and (c) 5 m.
Module 3-2 Unit Vectors, Adding Vectors by Components
•8 A person walks in the following pattern: 3.1 km north, then
2.4 km west, and finally 5.2 km south. (a) Sketch the vector dia-
gram that represents this motion. (b) How far and (c) in what di-
rection would a bird fly in a straight line from the same starting
point to the same final point?
•9 Two vectors are given by
and .
In unit-vector notation, find (a) , (b) , and (c) a third
vector such that .
•10 Find the (a) x, (b) y, and (c) zcomponents of the sum of
the displacements and whose components in meters are
cx7.4, cy3.8, cz6.1; dx4.4, dy2.0, dz3.3.
•11 (a) In unit-vector notation, what is the sum if
(4.0 m) (3.0 m) and ( 13.0 m) (7.0 m) ? What
are the (b) magnitude and (c) direction of ?a
:b
:j
ˆ
i
ˆ
b
:
j
ˆ
i
ˆ
a
:
a
:b
:
SSM
d
:
c
:
r
:
a
:b
:c
:0c
:
a
:b
:
a
:b
:
b
:(1.0 m)i
ˆ(1.0 m)j
ˆ(4.0 m)k
ˆ
a
:(4.0 m)i
ˆ(3.0 m)j
ˆ(1.0 m)k
ˆ
A
:
A
:
A
:
SSM
r
:
a
:
SSM
57
PROBLEMS
θ
d
Figure 3-27 Problem 6.
•12 A car is driven east for a distance of 50 km, then north for 30
km, and then in a direction 30° east of north for 25 km. Sketch the
vector diagram and determine (a) the magnitude and (b) the angle
of the car’s total displacement from its starting point.
•13 A person desires to reach a point that is 3.40 km from her
present location and in a direction that is 35.0° north of east.
However, she must travel along streets that are oriented either
north–south or east–west. What is the minimum distance she
could travel to reach her destination?
•14 You are to make four straight-line moves over a flat desert
floor, starting at the origin of an xy coordinate system and ending
at the xy coordinates (140 m, 30 m). The xcomponent and y
component of your moves are the following, respectively, in me-
ters: (20 and 60), then (bxand 70), then (20 and cy), then (60
and 70). What are (a) component bxand (b) component cy?
What are (c) the magnitude and (d) the angle (relative to the pos-
itive direction of the xaxis) of the overall displacement?
•15 The two vec-
WWWILWSSM
θ
x
y
r
Figure 3-26
Problem 2.
θ
Ox
y
2
θ
1
a
b
Figure 3-28 Problem 15.
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
•••32 In Fig. 3-31, a cube of edge
length asits with one corner at the ori-
gin of an xyz coordinate system. A
body diagonal is a line that extends
from one corner to another through
the center. In unit-vector notation,
what is the body diagonal that extends
from the corner at (a) coordinates (0,
0, 0), (b) coordinates (a, 0, 0), (c) coor-
dinates (0, a, 0), and (d) coordinates (a,a, 0)? (e) Determine the
CHAPTER 3 VECTORS
58
••30 Here are two vectors:
What are (a) the magnitude and (b) the angle (relative to ) of ?
What are (c) the magnitude and (d) the angle of ? What are (e)
the magnitude and (f) the angle of (g) the magnitude and
(h) the angle of ; and (i) the magnitude and (j) the angle of
? (k) What is the angle between the directions of
and ?
••31 In Fig. 3-30, a vector with a magnitude of 17.0 m is
directed at angle 56.0° counterclockwise from the axis.
What are the components (a) axand (b) ayof the vector? A sec-
ond coordinate system is inclined by angle 18.0° with respect
to the first. What are the components (c) and (d) in this
primed coordinate system?
a
y
a
x
x
a
:
a
:b
:b
:a
:
a
:b
:b
:a
:
a
:b
:;
b
:a
:
i
ˆ
a
:(4.0 m)i
ˆ(3.0 m)j
ˆ and b
:(6.0 m)i
ˆ(8.0 m)j
ˆ.
••20 An explorer is caught in a whiteout (in which the
snowfall is so thick that the ground cannot be distinguished from
the sky) while returning to base camp. He was supposed to travel
due north for 5.6 km, but when the snow clears, he discovers that
he actually traveled 7.8 km at 50° north of due east. (a) How far
and (b) in what direction must he now travel to reach base camp?
••21 An ant, crazed by the Sun on a hot Texas afternoon, darts
over an xy plane scratched in the dirt. The xand ycomponents of
four consecutive darts are the following, all in centimeters: (30.0,
40.0), (bx,70.0), (20.0, cy), (80.0, 70.0). The overall displace-
ment of the four darts has the xy components (140, 20.0).What
are (a) bxand (b) cy? What are the (c) magnitude and (d) angle
(relative to the positive direction of the xaxis) of the overall
displacement?
••22 (a) What is the sum of the following four vectors in unit-
vector notation? For that sum, what are (b) the magnitude, (c) the
angle in degrees, and (d) the angle in radians?
••23 If is added to , the result is a vector in the
positive direction of the yaxis, with a magnitude equal to that of .
What is the magnitude of ?
••24 Vector , which is directed along an xaxis, is to be addedA
:
B
:C
:
C
:3.0i
ˆ4.0j
ˆ
B
:
G
:: 4.00 m at 1.20 rad H
:: 6.00 m at 210
E
:
: 6.00 m at 0.900 rad F
:
: 5.00 m at 75.0
e
ab
c
h
i
y
x
j
k
l
m
n
o
p
q
r
s
t
u
w
v
d
fg
A
B
Figure 3-29 Problem 29.
y
x
z
a
a
a
Figure 3-31 Problem 32.
Ox
y
y'
x'
a'
x
ay
θ
θ
ax
a'
y
a
'
θ
'
Figure 3-30 Problem 31.
an ant’s displacement from the nest (find it in the figure) if the
ant enters the trail at point A? What are the (c) magnitude and
(d) angle if it enters at point B?
to vector , which has a magnitude of 7.0 m.The sum is a third vec-
tor that is directed along the yaxis, with a magnitude that is 3.0
times that of .What is that magnitude of ?
••25 Oasis Bis 25 km due east of oasis A. Starting from oasis
A, a camel walks 24 km in a direction 15° south of east and then
walks 8.0 km due north. How far is the camel then from oasis B?
••26 What is the sum of the following four vectors in (a) unit-
vector notation, and as (b) a magnitude and (c) an angle?
••27 If and thend
:
32i
ˆ4j
ˆ,d
:
1d
:
25d
:
3,d
:
1d
:
23d
:
3,
C
:(4.00 m)i
ˆ(6.00 m)j
ˆ D:
:5.00 m, at 235
A
:(2.00 m)i
ˆ(3.00 m)j
ˆ B:
:4.00 m, at 65.0
A
:
A
:
B
:
what are, in unit-vector notation, (a) and (b)
••28 Two beetles run across flat sand, starting at the same point.
Beetle 1 runs 0.50 m due east, then 0.80 m at 30° north of due east.
Beetle 2 also makes two runs; the first is 1.6 m at 40° east of due
north. What must be (a) the magnitude and (b) the direction of its
second run if it is to end up at the new location of beetle 1?
••29 Typical backyard ants often create a network of
chemical trails for guidance. Extending outward from the nest, a
trail branches (bifurcates) repeatedly, with 60° between the
branches. If a roaming ant chances upon a trail, it can tell the
way to the nest at any branch point: If it is moving away from
the nest, it has two choices of path requiring a small turn in
its travel direction, either 30° leftward or 30° rightward. If
it is moving toward the nest, it has only one such choice.
Figure 3-29 shows a typical ant trail, with lettered straight sec-
tions of 2.0 cm length and symmetric bifurcation of 60°. Path vis
parallel to the yaxis. What are the (a) magnitude and (b) angle
(relative to the positive direction of the superimposed xaxis) of
d
:
2?d
:
1
culate the angle between the two vectors given by
and .
••42 In a meeting of mimes, mime 1 goes through a displacement
and mime 2 goes through a displacement
. What are (a) , (b) ,
(c) , and (d) the com-
ponent of along the direction of
? (Hint: For (d), see Eq. 3-20 and
Fig. 3-18.)
••43 The three vectors in
Fig. 3-33 have magnitudes a3.00 m,
b4.00 m, and c10.0 m and angle
30.0°. What are (a) the xcompo-
nent and (b) the ycomponent of ;(c)
the xcomponent and (d) the ycom-
a
:
ILWSSM
d
:
2
d
:
1
(d
:
1d
:
2)d
:
2
d
:
1d
:
2
d
:
1d
:
2
d
:
2(3.0 m)i
ˆ(4.0 m)j
ˆ
d
:
1(4.0 m)i
ˆ(5.0 m)j
ˆ
b
:2.0i
ˆ1.0j
ˆ3.0k
ˆ
3.0j
ˆ3.0k
ˆa
:3.0i
ˆ
(a) , (b) , (c) , and(a
:b
:)b
:
a
:b
:
a
:b
:
is not shown.)
•34 Two vectors are presented as
and . Findb
:2.0i
ˆ4.0j
ˆ
a
:3.0i
ˆ5.0j
ˆ
59
PROBLEMS
angles that the body diagonals make with the adjacent edges.
(f) Determine the length of the body diagonals in terms of a.
Module 3-3 Multiplying Vectors
•33 For the vectors in Fig. 3-32, with a4, b3, and c5, what
are (a) the magnitude and (b) the direction
of , (c) the magnitude and (d) the di-
rection of , and (e) the magnitude
and (f) the direction of ? (The zaxisb
:c
:
a
:c
:
a
:b
:
ponent of ;and (e) the xcomponent and (f) the ycomponent of ? If
,what are the values of (g) pand (h) q?
••44 In the product , take q2,
.
What then is in unit-vector notation if BxBy?
Additional Problems
45 Vectors and lie in an xy plane. has magnitude 8.00 and
angle 130°; has components Bx7.72 and By9.20. (a)
What is What is in (b) unit-vector notation and
(c) magnitude-angle notation with spherical coordinates (see
Fig. 3-34)? (d) What is the angle between the directions of and
(Hint: Think a bit before you resort to a calculation.)
What is in (e) unit-vector notation and (f) magnitude-
angle notation with spherical coordinates?
A
:3.00k
ˆ
4A
:3B
:?
A
:
4A
:3B
:
5A
:B
:?
B
:A
:
B
:
A
:
B
:
v
:2.0i
ˆ4.0j
ˆ6.0k
ˆ and F
:4.0i
ˆ20j
ˆ12k
ˆ
F
:qv
:B
:
c
:pa
:qb
:c
:
b
:
θ
a
c
b
x
y
Figure 3-33 Problem 43.
φ
θ
y
x
z
Figure 3-34 Problem 45.
46 Vector has a magnitude of 5.0 m and is directed east.a
:
(d) the component of along the direc-
tion of . (Hint: For (d), consider Eq. 3-20
and Fig.3-18.)
•35 Two vectors, and ,lie in the xy plane.Their magnitudes are
4.50 and 7.30 units, respectively, and their directions are 320° and
85.0°, respectively, as measured counterclockwise from the positive
xaxis.What are the values of (a) and (b) ?
•36 If and , then what is
?
•37 Three vectors are given by
and . Find (a)
, (b) , and (c) .
••38 For the following three vectors, what is ?
••39 Vector has a magnitude of 6.00 units, vector has a mag-B
:
A
:
B
:3.00i
ˆ4.00j
ˆ2.00k
ˆ C
:7.00i
ˆ8.00j
ˆ
A
:2.00i
ˆ3.00j
ˆ4.00k
ˆ
3C
:(2A
:B
:)
a
:(b
:c
:)a
:(b
:c
:)a
:(b
:c
:)
c
:2.0i
ˆ2.0j
ˆ1.0k
ˆ
b
:1.0i
ˆ4.0j
ˆ2.0k
ˆ,
a
:3.0i
ˆ3.0j
ˆ2.0k
ˆ,
(d
:
1d
:
2)(d
:
14d
:
2)
d
:
25i
ˆ2j
ˆk
ˆ
d
:
13i
ˆ2j
ˆ4k
ˆ
r
:s
:
r
:s
:
s
:
r
:
b
:a
:
a
cb
y
x
Figure 3-32
Problems 33 and 54.
nitude of 7.00 units, and has a value of 14.0. What is the angle
between the directions of and ?
••40 Displacement is in the yz plane 63.0° from the positive
direction of the yaxis, has a positive zcomponent, and has a mag-
nitude of 4.50 m. Displacement is in the xz plane 30.0° from the
positive direction of the xaxis, has a positive zcomponent, and has
magnitude 1.40 m.What are (a) , (b) , and (c) the an-
gle between and ?
••41 Use the definition of scalar product,
, and the fact that to cal-a
:b
:axbxaybyazbz
a
:b
:ab cos
WWWILWSSM
d
:
2
d
:
1
d
:
1d
:
2
d
:
1d
:
2
d
:
2
d
:
1
B
:
A
:
A
:B
:
ax3.2, ay1.6, bx0.50, by4.5. (a) Find the angle between
the directions of and .There are two vectors in the xy plane that
are perpendicular to and have a magnitude of 5.0 m. One, vector
, has a positive xcomponent and the other, vector , a negative x
component. What are (b) the xcomponent and (c) the ycompo-
nent of vector , and (d) the xcomponent and (e) the ycomponent
of vector ?
49 A sailboat sets out from the U.S. side of Lake Erie for a
point on the Canadian side, 90.0 km due north. The sailor, how-
ever, ends up 50.0 km due east of the starting point. (a) How far
and (b) in what direction must the sailor now sail to reach the orig-
inal destination?
50 Vector is in the negative direction of a yaxis, and vector
is in the positive direction of an xaxis. What are the directions of
(a) and (b) What are the magnitudes of products (c)
and (d) What is the direction of the vector result-
ing from (e) and (f) ? What is the magnitude of the
vector product in (g) part (e) and (h) part (f)? What are the (i)
magnitude and (j) direction of ?d
:
1(d
:
2/4)
d
:
2d
:
1
d
:
1d
:
2
d
:
1(d
:
2/4)?d
:
1d
:
2
d
:
1/(4)?d
:
2/4
d
:
2
d
:
1
SSM
d
:c
:
d
:
c
:
a
:
b
:
a
:
Vector has a magnitude of 4.0 m and is directed 35° west of due
north.What are (a) the magnitude and (b) the direction of ?
What are (c) the magnitude and (d) the direction of ? (e)
Draw a vector diagram for each combination.
47 Vectors and lie in an xy plane. has magnitude 8.00
and angle 130°; has components Bx7.72 and By9.20.
What are the angles between the negative direction of the yaxis
and (a) the direction of , (b) the direction of the product
, and (c) the direction of ?
48 Two vectors and have the components, in meters,b
:
a
:
A
:(B
:3.00k
ˆ)A
:B
:A
:
B
:A
:
B
:
A
:
b
:a
:
a
:b
:
b
:
CHAPTER 3 VECTORS
60
51 Rock faults are ruptures along which opposite faces of rock
have slid past each other. In Fig. 3-35, points Aand Bcoincided be-
fore the rock in the foreground slid down to the right. The net dis-
placement is along the plane of the fault.The horizontal compo-
nent of is the strike-slip AC. The component of that is
directed down the plane of the fault is the dip-slip AD. (a) What is the
magnitude of the net displacement if the strike-slip is 22.0 m and
the dip-slip is 17.0 m? (b) If the plane of the fault is inclined at angle
52.0° to the horizontal, what is the vertical component of ?AB
9
:
AB
9
:
AB
9
:
AB
9
:
AB
9
:
58 A vector has a magnitude of 2.5 m and points north. What
are (a) the magnitude and (b) the direction of ? What are (c)
the magnitude and (d) the direction of ?
59 has the magnitude 12.0 m and is angled 60.0° counterclock-
wise from the positive direction of the xaxis of an xy coordinate
system. Also, on that same coordinate
system. We now rotate the system counterclockwise about the origin
by 20.0° to form an xysystem. On this new system, what are (a)
and (b) , both in unit-vector notation?
60 If and , then what are
(a) and (b) ?
61 (a) In unit-vector notation, what is if
5.0 4.0 6.0 , 2.0 2.0 3.0 , and 4.0
3.0 2.0 ? (b) Calculate the angle between and the positive z
axis. (c) What is the component of along the direction of ? (d)
What is the component of perpendicular to the direction of but
in the plane of and ? (Hint: For (c), see Eq. 3-20 and Fig. 3-18;
for (d), see Eq. 3-24.)
62 A golfer takes three putts to get the ball into the hole. The
first putt displaces the ball 3.66 m north, the second 1.83 m south-
east, and the third 0.91 m southwest. What are (a) the magnitude
and (b) the direction of the displacement needed to get the ball
into the hole on the first putt?
63 Here are three vectors in meters:
What results from (a) (b) and
(c) ?
64 A room has dimensions 3.00 m (height)
3.70 m 4.30 m. A fly starting at one corner flies around, ending
up at the diagonally opposite corner. (a) What is the magnitude of
its displacement? (b) Could the length of its path be less than this
magnitude? (c) Greater? (d) Equal? (e) Choose a suitable coordi-
nate system and express the components of the displacement vec-
tor in that system in unit-vector notation. (f) If the fly walks, what
is the length of the shortest path? (Hint: This can be answered
without calculus. The room is like a box. Unfold its walls to flatten
them into a plane.)
65 A protester carries his sign of protest, starting from the ori-
gin of an xyz coordinate system, with the xy plane horizontal. He
moves 40 m in the negative direction of the xaxis, then 20 m
along a perpendicular path to his left, and then 25 m up a water
tower. (a) In unit-vector notation, what is the displacement of
the sign from start to end? (b) The sign then falls to the foot of
the tower.What is the magnitude of the displacement of the sign
from start to this new end?
66 Consider in the positive direction of x, in the positive di-
rection of y, and a scalar d. What is the direction of if dis
(a) positive and (b) negative? What is the magnitude of (c)
and (d) ? What is the direction of the vector resulting from
(e) and (f) ? (g) What is the magnitude of the vector
product in (e)? (h) What is the magnitude of the vector product in
(f)? What are (i) the magnitude and (j) the direction of if d
is positive?
b
:/da
:
b
:a
:
a
:b
:
a
:b
:/d
a
:b
:
b
:/d
b
:
a
:
WWWSSM
d
:
1(d
:
2d
:
3)
d
:
1(d
:
2d
:
3),d
:
1(d
:
2d
:
3),
d
:
32.0i
ˆ3.0j
ˆ1.0k
ˆ.
d
:
22.0i
ˆ4.0j
ˆ2.0k
ˆ
d
:
13.0i
ˆ3.0j
ˆ2.0k
ˆ
b
:
a
:
b
:
a
:
b
:
a
:
r
:
k
ˆ
j
ˆ
i
ˆc
:k
ˆ
j
ˆi
ˆb
:k
ˆ
j
ˆi
ˆa
:
c
:
b
:a
:r
:
b
:
a
:
c
:3i
ˆ4j
ˆ
a
:b
:2c
:,a
:b
:4c
:,
B
:A
:
B
:(12.0 m)i
ˆ(8.00 m)j
ˆ
A
:
3.0d
:4.0d
:
d
:
A
D
C
Strike-slip
Dip-slip
Fault plane
B
φ
Figure 3-35 Problem 51.
52 Here are three displacements, each measured in meters:
and
. (a) What is ? (b) What is the
angle between and the positive zaxis? (c) What is the compo-
nent of along the direction of (d) What is the component of
that is perpendicular to the direction of and in the plane of
and (Hint: For (c), consider Eq. 3-20 and Fig. 3-18; for (d), con-
sider Eq. 3-24.)
53 A vector of magnitude 10 units and another vector
of magnitude 6.0 units differ in directions by 60°. Find (a) the
scalar product of the two vectors and (b) the magnitude of the vec-
tor product .
54 For the vectors in Fig. 3-32, with a4, b3, and c5, calcu-
late (a) , (b) , and (c) .
55 A particle undergoes three successive displacements in a
plane, as follows: 4.00 m southwest; then 5.00 m east; and
finally 6.00 m in a direction 60.0° north of east. Choose a coor-
dinate system with the yaxis pointing north and the xaxis pointing
east.What are (a) the xcomponent and (b) the ycomponent of ?
What are (c) the xcomponent and (d) the ycomponent of ?
What are (e) the xcomponent and (f) the ycomponent of ?
Next, consider the net displacement of the particle for the three
successive displacements. What are (g) the xcomponent, (h) the y
component, (i) the magnitude, and ( j) the direction of the net dis-
placement? If the particle is to return directly to the starting point,
(k) how far and (l) in what direction should it move?
56 Find the sum of the following four vectors in (a) unit-vector
notation, and as (b) a magnitude and (c) an angle relative to x.
: 10.0 m, at 25.0° counterclockwise from x
: 12.0 m, at 10.0° counterclockwise from y
: 8.00 m, at 20.0° clockwise from y
: 9.00 m, at 40.0° counterclockwise from y
57 If is added to , the result is 6.0 1.0 . If is subtracted
from ,the result is 4.0 7.0 .What is the magnitude of ?A
:
j
ˆ
i
ˆ
A
:B
:
j
ˆ
i
ˆ
A
:
B
:
SSM
S
:
R
:
Q
:
P
:
d
:
3
d
:
2
d
:
1
d
:
3,
d
:
2,d
:
1,
b
:c
:
a
:c
:
a
:b
:
a
:b
:
b
:
a
:
SSM
d
:
2?
d
:
1
d
:
2
d
:
1
d
:
2?d
:
1
r
:
r
:d
:
1d
:
2d
:
3
4.0i
ˆ3.0j
ˆ2.0k
ˆd
:
3d
:
21.0i
ˆ2.0j
ˆ3.0k
ˆ,d
:
14.0i
ˆ5.0j
ˆ6.0k
ˆ,
61
PROBLEMS
67 Let be directed to the east, be directed to the north, and k
ˆ
j
ˆ
i
ˆ72 A fire ant, searching for hot sauce in a picnic area, goes
through three displacements along level ground: lfor 0.40 m
southwest (that is, at 45° from directly south and from directly
west), 2for 0.50 m due east, 3for 0.60 m at 60° north of east.
Let the positive xdirection be east and the positive ydirection
be north. What are (a) the xcomponent and (b) the ycompo-
nent of l? Next, what are (c) the xcomponent and (d) the y
component of 2? Also, what are (e) the xcomponent and (f)
the ycomponent of 3?
What are (g) the xcomponent, (h) the ycomponent, (i) the
magnitude, and (j) the direction of the ant’s net displacement? If
the ant is to return directly to the starting point, (k) how far and (1)
in what direction should it move?
73 Two vectors are given by 3.0 5.0 and 2.0 4.0 .j
ˆ
i
ˆ
b
:
j
ˆ
i
ˆ
a
:
d
:
d
:
d
:
d
:
d
:
d
:
a
b
f
Figure 3-38 Problem 79.
BOSTON
and Vicinity
Wellesley
Waltham
Brookline
Newton
Arlington
Lexington Woburn
Medford
Lynn
Salem
Quincy
5 10 km
BOSTON Massachusetts
Bay
Bank
Walpole
Framingham
Weymouth
Dedham
Winthrop
N
Figure 3-36 Problem 68.
be directed upward. What are the values of products (a) , (b)
()(), and (c) ()? What are the directions (such as eastj
ˆ
j
ˆ
j
ˆ
k
ˆk
ˆ
i
ˆ
or down) of products (d) ,(e) ()(), and (f) () ()?
68 A bank in downtown Boston is robbed (see the map in
Fig. 3-36). To elude police, the robbers escape by helicopter, mak-
ing three successive flights described by the following displace-
ments: 32 km, 45° south of east; 53 km, 26° north of west; 26 km, 18°
east of south. At the end of the third flight they are captured. In
what town are they apprehended?
j
ˆ
k
ˆ
j
ˆ
i
ˆ
j
ˆ
k
ˆ
69 A wheel with a radius of 45.0 cm
rolls without slipping along a hori-
zontal floor (Fig. 3-37). At time t1,
the dot Ppainted on the rim of the
wheel is at the point of contact be-
tween the wheel and the floor.At a
later time t2, the wheel has rolled
through one-half of a revolution.
What are (a) the magnitude and (b)
the angle (relative to the floor) of
the displacement of P?
70 A woman walks 250 m in the direction 30° east of north, then
175 m directly east. Find (a) the magnitude and (b) the angle of her
final displacement from the starting point. (c) Find the distance she
walks. (d) Which is greater, that distance or the magnitude of her
displacement?
71 A vector has a magnitude 3.0 m and is directed south. What
are (a) the magnitude and (b) the direction of the vector 5.0 ? What
are (c) the magnitude and (d) the direction of the vector 2.0 ?d
:
d
:
d
:
P
At time t1At time t
2
P
Figure 3-37 Problem 69.
Find (a) , (b) , (c) , and (d) the component of
along the direction of .
74 Vector lies in the yz plane 63.0from the positive direction
of the yaxis, has a positive zcomponent, and has magnitude 3.20
units. Vector lies in the xz plane 48.0from the positive direction
of the xaxis, has a positive zcomponent, and has magnitude 1.40
units. Find (a) , (b) , and (c) the angle between and .
75 Find (a) “north cross west,” (b) “down dot south,” (c) “east
cross up,” (d) “west dot west,” and (e) “south cross south.” Let each
“vector” have unit magnitude.
76 A vector , with a magnitude of 8.0 m, is added to a vector ,
which lies along an xaxis. The sum of these two vectors is a third
vector that lies along the yaxis and has a magnitude that is twice
the magnitude of .What is the magnitude of ?
77 A man goes for a walk, starting from the origin of an xyz
coordinate system, with the xy plane horizontal and the xaxis east-
ward. Carrying a bad penny, he walks 1300 m east, 2200 m north,
and then drops the penny from a cliff 410 m high. (a) In unit-vector
notation, what is the displacement of the penny from start to its
landing point? (b) When the man returns to the origin, what is the
magnitude of his displacement for the return trip?
78 What is the magnitude of () if a3.90, b2.70,
and the angle between the two vectors is 63.0°?
79 In Fig. 3-38, the magnitude of is 4.3, the magnitude of is
5.4, and
46°. Find the area of the triangle contained between
the two vectors and the thin diagonal line.
b
:
a
:
a
:
b
:
a
:
A
:
A
:
A
:
B
:
b
:
a
:
b
:
a
:
b
:
a
:
b
:
a
:
b
:
a
:
b
:
(a
:b
:)a
:b
:
b
:
a
:
62
CHAPTER 4
Motion in Two and Three Dimensions
4-1 POSITION AND DISPLACEMENT
Learning Objectives
After reading this module, you should be able to . . .
4.01 Draw two-dimensional and three-dimensional position
vectors for a particle, indicating the components along the
axes of a coordinate system.
4.02 On a coordinate system, determine the direction and
magnitude of a particle’s position vector from its compo-
nents, and vice versa.
4.03 Apply the relationship between a particle’s displace-
ment vector and its initial and final position vectors.
Key Ideas
●The location of a particle relative to the origin of a coordi-
nate system is given by a position vector , which in unit-
vector notation is
Here x,y, and zare the vector components of position
vector , and and zare its scalar components (as well
as the coordinates of the particle).
●A position vector is described either by a magnitude and
x,y,r
:
k
ˆ
j
ˆ
i
ˆ
r
:xi
ˆyj
ˆzk
ˆ.
r
:
one or two angles for orientation, or by its vector or scalar
components.
●If a particle moves so that its position vector changes from
to , the particle’s displacement is
The displacement can also be written as
xi
ˆyj
ˆzk
ˆ.
r
:(x2x1)i
ˆ(y2y1)j
ˆ(z2z1)k
ˆ
r
:r
:
2r
:
1.
r
:
r
:
2
r
:
1
What Is Physics?
In this chapter we continue looking at the aspect of physics that analyzes
motion, but now the motion can be in two or three dimensions. For example,
medical researchers and aeronautical engineers might concentrate on the
physics of the two- and three-dimensional turns taken by fighter pilots in dog-
fights because a modern high-performance jet can take a tight turn so quickly
that the pilot immediately loses consciousness. A sports engineer might focus
on the physics of basketball. For example, in a free throw (where a player gets
an uncontested shot at the basket from about 4.3 m), a player might employ the
overhand push shot, in which the ball is pushed away from about shoulder
height and then released. Or the player might use an underhand loop shot, in
which the ball is brought upward from about the belt-line level and released.
The first technique is the overwhelming choice among professional players, but
the legendary Rick Barry set the record for free-throw shooting with the under-
hand technique.
Motion in three dimensions is not easy to understand. For example, you are
probably good at driving a car along a freeway (one-dimensional motion) but
would probably have a difficult time in landing an airplane on a runway (three-
dimensional motion) without a lot of training.
In our study of two- and three-dimensional motion, we start with position
and displacement.
Position and Displacement
One general way of locating a particle (or particle-like object) is with a position
vector , which is a vector that extends from a reference point (usually the
origin) to the particle. In the unit-vector notation of Module 3-2, can be written
(4-1)
where x,y, and zare the vector components of and the coefficients x,y, and z
are its scalar components.
The coefficients x,y, and zgive the particle’s location along the coordinate
axes and relative to the origin; that is, the particle has the rectangular coordinates
(x,y,z). For instance,Fig. 4-1 shows a particle with position vector
and rectangular coordinates (3 m, 2 m, 5 m). Along the xaxis the particle is
3 m from the origin, in the direction. Along the yaxis it is 2 m from the
origin, in the direction. Along the zaxis it is 5 m from the origin, in the
direction.
As a particle moves, its position vector changes in such a way that the vector
always extends to the particle from the reference point (the origin). If the posi-
tion vector changes—say, from to during a certain time interval—then the
particle’s displacement during that time interval is
(4-2)
Using the unit-vector notation of Eq. 4-1, we can rewrite this displacement as
or as (4-3)
where coordinates (x1,y1,z1) correspond to position vector and coordinates
(x2,y2,z2) correspond to position vector .We can also rewrite the displacement
by substituting xfor (x2x1), yfor (y2y1), and zfor (z2z1):
(4-4)r
:xi
ˆyj
ˆzk
ˆ.
r
:
2
r
:
1
r
:(x2x1)i
ˆ(y2y1)j
ˆ(z2z1)k
ˆ,
r
:(x2i
ˆy2j
ˆz2k
ˆ)(x1i
ˆy1j
ˆz1k
ˆ)
r
:r
:
2r
:
1.
r
:
r
:
2
r
:
1
k
ˆ
j
ˆi
ˆ
r
:(3 m)i
ˆ(2 m)j
ˆ(5 m)k
ˆ
r
:
k
ˆ
j
ˆ
i
ˆ
r
:xi
ˆyj
ˆzk
ˆ,
r
:
r
:
Figure 4-1 The position vector for a parti-
cle is the vector sum of its vector compo-
nents.
r
:
y
x
z
(–3 m)i
(2 m)j
(5 m)k
O
ˆ
ˆ
ˆ
r
To locate the
particle, this
is how far
parallel to z.
This is how far
parallel to y.
This is how far
parallel to x.
position vector . Let’s evaluate those coordinates at the
given time, and then we can use Eq. 3-6 to evaluate the mag-
nitude and orientation of the position vector.
r
:
Sample Problem 4.01 Two-dimensional position vector, rabbit run
A rabbit runs across a parking lot on which a set of
coordinate axes has, strangely enough, been drawn. The co-
ordinates (meters) of the rabbit’s position as functions of
time t(seconds) are given by
x0.31t27.2t28 (4-5)
and y0.22t29.1t30. (4-6)
(a) At t15 s, what is the rabbit’s position vector in unit-
vector notation and in magnitude-angle notation?
KEY IDEA
The xand ycoordinates of the rabbit’s position, as given by
Eqs. 4-5 and 4-6, are the scalar components of the rabbit’s
r
:
Calculations: We can write
(4-7)
(We write rather than because the components are
functions of t, and thus is also.)
At t15 s, the scalar components are
x(0.31)(15)2(7.2)(15) 28 66 m
and y(0.22)(15)2(9.1)(15) 30 57 m,
so (Answer)r
:(66 m)i
ˆ(57 m)j
ˆ,
r
:r
:
r
:(t)
r
:(t)x(t)i
ˆy(t)j
ˆ.
63
4-1 POSITION AND DISPLACEMENT
4-2 AVERAGE VELOCITY AND INSTANTANEOUS VELOCITY
64 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
x (m)
0
20
40
–20
–40
–60
y (m)
20 40 60 80
(b)25 s 20 s
15 s
10 s
5 s
t = 0 s
This is the path with
various times indicated.
x (m)
0
20
40
–20
–40
–60
y (m)
20 40 60 80
(a)
–41°
r
This is the ycomponent.
To locate the
rabbit, this is the
xcomponent.
Figure 4-2 (a) A rabbit’s position vector
at time t15 s. The scalar compo-
nents of are shown along the axes.
(b) The rabbit’s path and its position at
six values of t.
r
:
r
:
Additional examples, video, and practice available at WileyPLUS
Check: Although u139° has the same tangent as 41°,
the components of position vector indicate that the de-
sired angle is 139° 180° 41°.
(b) Graph the rabbit’s path for t0 to t25 s.
Graphing: We have located the rabbit at one instant, but to
see its path we need a graph. So we repeat part (a) for sev-
eral values of tand then plot the results. Figure 4-2bshows
the plots for six values of tand the path connecting them.
r
:
which is drawn in Fig. 4-2a. To get the magnitude and angle
of , notice that the components form the legs of a right tri-
angle and ris the hypotenuse. So, we use Eq. 3-6:
(Answer)
and . (Answer)
tan1y
xtan1
57 m
66 m
41
87 m,
r2x2y22(66 m)2(57 m)2
r
:
4.06 In magnitude-angle and unit-vector notations, relate a parti-
cle’s initial and final position vectors, the time interval between
those positions, and the particle’s average velocity vector.
4.07 Given a particle’s position vector as a function of time,
determine its (instantaneous) velocity vector.
Learning Objectives
After reading this module, you should be able to . . .
4.04 Identify that velocity is a vector quantity and thus has
both magnitude and direction and also has components.
4.05 Draw two-dimensional and three-dimensional velocity
vectors for a particle, indicating the components along the
axes of the coordinate system.
which can be rewritten in unit-vector notation as
where and
●The instantaneous velocity of a particle is always directed
along the tangent to the particle’s path at the particle’s
position.
v
:
vzdz/dt.vxdx/dt, vydy/dt,
v
:vxi
ˆvyj
ˆvzk
ˆ,
Key Ideas
●If a particle undergoes a displacement in time interval t,
its average velocity for that time interval is
●As tis shrunk to 0, reaches a limit called either the
velocity or the instantaneous velocity :
v
:dr
:
dt ,
v
:
v
:
avg
v
:
avg r
:
t.
v
:
avg
r
:
65
4-2 AVERAGE VELOCITY AND INSTANTANEOUS VELOCITY
Average Velocity and Instantaneous Velocity
If a particle moves from one point to another, we might need to know how fast it
moves. Just as in Chapter 2, we can define two quantities that deal with “how
fast”: average velocity and instantaneous velocity. However, here we must con-
sider these quantities as vectors and use vector notation.
If a particle moves through a displacement in a time interval t, then its
average velocity is
or (4-8)
This tells us that the direction of (the vector on the left side of Eq. 4-8) must
be the same as that of the displacement (the vector on the right side). Using
Eq. 4-4, we can write Eq. 4-8 in vector components as
(4-9)
For example, if a particle moves through displacement in
2.0 s, then its average velocity during that move is
That is, the average velocity (a vector quantity) has a component of 6.0 m/s along
the xaxis and a component of 1.5 m/s along the zaxis.
When we speak of the velocity of a particle, we usually mean the particle’s
instantaneous velocity at some instant. This is the value that approaches
in the limit as we shrink the time interval tto 0 about that instant. Using the lan-
guage of calculus, we may write as the derivative
(4-10)
Figure 4-3 shows the path of a particle that is restricted to the xy plane. As
the particle travels to the right along the curve, its position vector sweeps to the
right. During time interval t, the position vector changes from to and the
particle’s displacement is .
To find the instantaneous velocity of the particle at, say, instant t1(when the
particle is at position 1), we shrink interval tto 0 about t1. Three things happen
as we do so. (1) Position vector in Fig. 4-3 moves toward so that shrinksr
:
r
:
1
r
:
2
r
:
r
:
2
r
:
1
v
:dr
:
dt .
v
:
v
:
avg
v
:
v
:
v
:
avg r
:
t(12 m)i
ˆ(3.0 m)k
ˆ
2.0 s (6.0 m/s)i
ˆ(1.5 m/s)k
ˆ.
(12 m)i
ˆ(3.0 m)k
ˆ
v
:
avg xi
ˆyj
ˆzk
ˆ
tx
t i
ˆy
t j
ˆz
t k
ˆ.
r
:
v
:
avg
v
:
avg r
:
t.
average velocity displacement
time interval ,
v
:
avg
r
:
Figure 4-3 The displacement of a particle
during a time interval , from position 1 with
position vector at time t1to position 2
with position vector at time t2. The tangent
to the particle’s path at position 1 is shown.
r
:
2
r
:
1
t
r
:
r1r2
Path
Tangent
O
y
x
1
2
r
Δ
As the particle moves,
the position vector
must change.
This is the
displacement.
toward zero. (2) The direction of (and thus of ) approaches the
direction of the line tangent to the particle’s path at position 1. (3) The average
velocity approaches the instantaneous velocity at t1.v
:
v
:
avg
v
:
avg
r
:/t
Figure 4-4 The velocity of a
particle, along with the scalar
components of .
v
:
v
:
Path
O
y
x
Tangent
vy
vx
v
The velocity vector is always
tangent to the path.
These are the x and y
components of the vector
at this instant.
Checkpoint 1
The figure shows a circular path taken by a particle.
If the instantaneous velocity of the particle is
, through which quadrant is the par-
ticle moving at that instant if it is traveling (a) clockwise
and (b) counterclockwise around the circle? For both
cases, draw on the figure.
v
:
(2 m/s)i
ˆ(2 m/s)j
ˆv
:
y
x
66 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
The direction of the instantaneous velocity of a particle is always tangent to the
particle’s path at the particle’s position.
v
:
The result is the same in three dimensions: is always tangent to the particle’s path.
To write Eq. 4-10 in unit-vector form, we substitute for from Eq. 4-1:
This equation can be simplified somewhat by writing it as
(4-11)
where the scalar components of are
(4-12)
For example,dx/dt is the scalar component of along the xaxis.Thus, we can find
the scalar components of by differentiating the scalar components of .
Figure 4-4 shows a velocity vector and its scalar xand ycomponents. Note
that is tangent to the particle’s path at the particle’s position. Caution: When a
position vector is drawn, as in Figs. 4-1 through 4-3, it is an arrow that extends
from one point (a “here”) to another point (a “there”). However, when a velocity
vector is drawn, as in Fig. 4-4, it does not extend from one point to another.
Rather, it shows the instantaneous direction of travel of a particle at the tail, and
its length (representing the velocity magnitude) can be drawn to any scale.
v
:
v
:
r
:
v
:
v
:
vxdx
dt , vydy
dt , and vzdz
dt .
v
:
v
:vxi
ˆvyj
ˆvzk
ˆ,
v
:d
dt (xi
ˆyj
ˆzk
ˆ)dx
dt i
ˆdy
dt j
ˆdz
dt k
ˆ.
r
:
v
:
In the limit as , we have and, most important here, takes
on the direction of the tangent line.Thus, has that direction as well:v
:
v
:
avg
v
:
avg :v
:
t:0
67
4-3 AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION
(Answer)
and
(Answer)
Check: Is the angle 130° or 130° 180° 50°?
tan11.19 130.
tan1vy
vx
tan1
2.5 m/s
2.1 m/s
3.3 m/s
v2vx
2vy
22(2.1 m/s)2(2.5 m/s)2
For the rabbit in the preceding sample problem, find the
velocity at time t15 s.
KEY IDEA
We can find by taking derivatives of the components of
the rabbit’s position vector.
Calculations: Applying the vxpart of Eq. 4-12 to Eq. 4-5,
we find the xcomponent of to be
(4-13)
At t15 s, this gives vx2.1 m/s. Similarly, applying the
vypart of Eq. 4-12 to Eq. 4-6, we find
(4-14)
At t15 s, this gives vy2.5 m/s. Equation 4-11 then yields
(Answer)
which is shown in Fig. 4-5, tangent to the rabbit’s path and in
the direction the rabbit is running at t15 s.
To get the magnitude and angle of , either we use a
vector-capable calculator or we follow Eq. 3-6 to write
v
:
v
:(2.1 m/s)i
ˆ(2.5 m/s)j
ˆ,
0.44t9.1.
vydy
dt d
dt (0.22t29.1t30)
0.62t7.2.
vxdx
dt d
dt (0.31t27.2t28)
v
:
v
:
v
:
Sample Problem 4.02 Two-dimensional velocity, rabbit run
Additional examples, video, and practice available at WileyPLUS
Figure 4-5 The rabbit’s velocity at t15 s.v
:
–130°
x (m)
0
20
40
–20
–40
–60
y (m)
20 40 60 80
x
v
These are the x and y
components of the vector
at this instant.
4-3 AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION
the average acceleration vector in magnitude-angle and
unit-vector notations.
4.11 Given a particle’s velocity vector as a function of time,
determine its (instantaneous) acceleration vector.
4.12 For each dimension of motion, apply the constant-
acceleration equations (Chapter 2) to relate acceleration,
velocity, position, and time.
Learning Objectives
After reading this module, you should be able to . . .
4.08 Identify that acceleration is a vector quantity and thus has
both magnitude and direction and also has components.
4.09 Draw two-dimensional and three-dimensional accelera-
tion vectors for a particle, indicating the components.
4.10 Given the initial and final velocity vectors of a particle
and the time interval between those velocities, determine
either the acceleration or the instantaneous acceleration :
●In unit-vector notation,
where and azdvz/dt.axdvx/dt, aydvy/dt,
a
:axi
ˆayj
ˆazk
ˆ,
a
:dv
:
dt .
a
:
Key Ideas
●If a particle’s velocity changes from to in time interval
t, its average acceleration during tis
●As tis shrunk to 0, reaches a limiting value called a
:
avg
a
:
avg v
:
2v
:
1
tv
:
t.
v
:
2
v
:
1
68 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Average Acceleration and Instantaneous Acceleration
When a particle’s velocity changes from to in a time interval t, its average
acceleration during tis
or (4-15)
If we shrink tto zero about some instant, then in the limit approaches the
instantaneous acceleration (or acceleration)at that instant; that is,
(4-16)
If the velocity changes in either magnitude or direction (or both), the particle
must have an acceleration.
We can write Eq. 4-16 in unit-vector form by substituting Eq. 4-11 for to obtain
We can rewrite this as
(4-17)
where the scalar components of are
(4-18)
To find the scalar components of , we differentiate the scalar components of .
Figure 4-6 shows an acceleration vector and its scalar components for a
particle moving in two dimensions. Caution: When an acceleration vector is
drawn, as in Fig. 4-6, it does not extend from one position to another. Rather, it
shows the direction of acceleration for a particle located at its tail, and its length
(representing the acceleration magnitude) can be drawn to any scale.
a
:
v
:
a
:
axdvx
dt , aydvy
dt , and azdvz
dt .
a
:
a
:axi
ˆayj
ˆazk
ˆ,
dvx
dt i
ˆdvy
dt j
ˆdvz
dt k
ˆ.
a
:d
dt (vxi
ˆvyj
ˆvzk
ˆ)
v
:
a
:dv
:
dt .
a
:
a
:
avg
a
:
avg v
:
2v
:
1
tv
:
t.
average
acceleration change in velocity
time interval ,
a
:
avg
v
:
2
v
:
1
O
y
x
ay
ax
Path
a
These are the x and y
components of the vector
at this instant.
Figure 4-6 The acceleration of a particle and the
scalar components of .
a
:
a
:
69
4-3 AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION
Sample Problem 4.03 Two-dimensional acceleration, rabbit run
Additional examples, video, and practice available at WileyPLUS
For the rabbit in the preceding two sample problems, find
the acceleration at time t15 s.
KEY IDEA
We can find by taking derivatives of the rabbit’s velocity
components.
Calculations: Applying the axpart of Eq. 4-18 to Eq. 4-13,
we find the xcomponent of to be
Similarly, applying the aypart of Eq. 4-18 to Eq. 4-14 yields
the ycomponent as
We see that the acceleration does not vary with time (it is a
constant) because the time variable tdoes not appear in the
expression for either acceleration component. Equation 4-17
then yields
(Answer)
which is superimposed on the rabbit’s path in Fig. 4-7.
To get the magnitude and angle of , either we use a
vector-capable calculator or we follow Eq. 3-6. For the mag-
nitude we have
(Answer)
For the angle we have
However, this angle, which is the one displayed on a calcula-
tor, indicates that is directed to the right and downward in
Fig. 4-7. Yet, we know from the components that must be
directed to the left and upward. To find the other angle that
a
:
a
:
tan1ay
ax
tan1
0.44 m/s2
0.62 m/s2
35.
0.76 m/s2.
a2ax
2ay
22(0.62 m/s2)2(0.44 m/s2)2
a
:
a
:(0.62 m/s2)i
ˆ(0.44 m/s2)j
ˆ,
aydvy
dt d
dt (0.44t9.1) 0.44 m/s2.
axdvx
dt d
dt (0.62t7.2) 0.62 m/s2.
a
:
a
:
a
:
x (m)
0
20
40
–20
–40
–60
y (m)
20 40 60 80
x
145°
a
These are the x and y
components of the vector
at this instant.
Figure 4-7 The acceleration of the rabbit at t15 s. The rabbit
happens to have this same acceleration at all points on its path.
a
:
has the same tangent as 35° but is not displayed on a cal-
culator,we add 180°:
35° 180° 145°. (Answer)
This is consistent with the components of because it gives
a vector that is to the left and upward. Note that has the
same magnitude and direction throughout the rabbit’s run
because the acceleration is constant. That means that
we could draw the very same vector at any other point
along the rabbit’s path (just shift the vector to put its tail at
some other point on the path without changing the length
or orientation).
This has been the second sample problem in which we
needed to take the derivative of a vector that is written in
unit-vector notation. One common error is to neglect the unit
vectors themselves, with a result of only a set of numbers and
symbols. Keep in mind that a derivative of a vector is always
another vector.
a
:
a
:
Checkpoint 2
Here are four descriptions of the position (in meters) of a puck as it moves in an xy plane:
(1) x3t24t2 and y6t24t(3)
(2) x3t34tand y5t26 (4)
Are the xand yacceleration components constant? Is acceleration constant?a
:
r
:(4t32t)i
ˆ3j
ˆ
r
:2t2i
ˆ(4t3)j
ˆ
70 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Projectile Motion
We next consider a special case of two-dimensional motion:A particle moves in a
vertical plane with some initial velocity but its acceleration is always the free-
fall acceleration , which is downward. Such a particle is called a projectile (mean-
ing that it is projected or launched), and its motion is called projectile motion. A
projectile might be a tennis ball (Fig. 4-8) or baseball in flight, but it is not a duck
in flight. Many sports involve the study of the projectile motion of a ball. For ex-
ample, the racquetball player who discovered the Z-shot in the 1970s easily won
his games because of the ball’s perplexing flight to the rear of the court.
Our goal here is to analyze projectile motion using the tools for two-
dimensional motion described in Module 4-1 through 4-3 and making the
assumption that air has no effect on the projectile. Figure 4-9, which we shall ana-
lyze soon, shows the path followed by a projectile when the air has no effect.The
projectile is launched with an initial velocity that can be written as
(4-19)
The components v0xand v0ycan then be found if we know the angle u0between
and the positive xdirection:
v0xv0cos u0and v0yv0sin u0. (4-20)
During its two-dimensional motion, the projectile’s position vector and velocity
vector change continuously, but its acceleration vector is constant and always
directed vertically downward.The projectile has no horizontal acceleration.
Projectile motion, like that in Figs. 4-8 and 4-9, looks complicated, but we
have the following simplifying feature (known from experiment):
a
:
v
:
r
:
v
:
0
v
:
0v0xi
ˆv0yj
ˆ.
v
:
0
g
:
v
:
0
Figure 4-8 A stroboscopic photograph of
a yellow tennis ball bouncing off a hard
surface. Between impacts, the ball has
projectile motion.
Richard Megna/Fundamental Photographs
●In projectile motion, a particle is launched into the air with a
speed v0and at an angle u0(as measured from a horizontal x
axis). During flight, its horizontal acceleration is zero and its
vertical acceleration is g(downward on a vertical yaxis).
●The equations of motion for the particle (while in flight) can
be written as
v2
y(v0sin
0)22g(yy0).
vyv0sin
0gt,
yy0(v0 sin
0)t1
2gt2,
xx0(v0cos
0)t,
●The trajectory (path) of a particle in projectile motion is par-
abolic and is given by
if x0and y0are zero.
●The particle’s horizontal range R, which is the horizontal
distance from the launch point to the point at which the parti-
cle returns to the launch height, is
Rv0
2
gsin 2
0.
y(tan
0)xgx2
2(v0cos
0)2,
4-4 PROJECTILE MOTION
4.14 Given the launch velocity in either magnitude-angle or
unit-vector notation, calculate the particle’s position, dis-
placement, and velocity at a given instant during the flight.
4.15 Given data for an instant during the flight, calculate the
launch velocity.
Learning Objectives
After reading this module, you should be able to . . .
4.13 On a sketch of the path taken in projectile motion,
explain the magnitudes and directions of the velocity
and acceleration components during the flight.
Key Ideas
In projectile motion, the horizontal motion and the vertical motion are indepen-
dent of each other; that is, neither motion affects the other.
71
4-4 PROJECTILE MOTION
A
x
y
O
0
θ
v0
v0y
v0x
y
O
x
O
vx
vyv
x
y
y
O
vy= 0
v
x
y
O
vx
vyv
x
y
O
vx
x
Ov0x
x
Ovx
x
O
vx
x
O
vx
x
O
vx
y
O
v0y
O
vy
y
y
O
vy= 0
y
O
vy
y
vx
vy
θ
v
vx
vy
Vertical motion
Vertical velocity
Launch Launch
Speed decreasing
Constant velocity
Speed increasing
Constant velocity
Constant velocity
Stopped at
maximum
height
Constant velocity
+➡Horizontal motion Projectile motion
Launch velocity
Launch angle
This vertical motion plus
this horizontal motion
produces this projectile motion.
Figure 4-9 The projectile motion of an object launched into the air at the origin of a coordinate system and with launch
velocity at angle u0.The motion is a combination of vertical motion (constant acceleration) and horizontal motion
(constant velocity), as shown by the velocity components.
v
:
0
72 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Checkpoint 3
At a certain instant, a fly ball has velocity (the xaxis is horizontal, the
yaxis is upward, and is in meters per second). Has the ball passed its highest point?v
:
v
:25i
ˆ4.9j
ˆ
Figure 4-11 The projectile ball always
hits the falling can. Each falls a distance h
from where it would be were there no
free-fall acceleration.
M
Can
h
Zero-gpath
G
The ball and the can fall
the same distance h.
Figure 4-10 One ball is released from rest at
the same instant that another ball is shot
horizontally to the right. Their vertical
motions are identical.
Richard Megna/Fundamental Photographs
The Horizontal Motion
Now we are ready to analyze projectile motion, horizontally and vertically.
We start with the horizontal motion. Because there is no acceleration in the hori-
zontal direction, the horizontal component vxof the projectile’s velocity remains
unchanged from its initial value v0xthroughout the motion, as demonstrated in
Fig. 4-12. At any time t, the projectile’s horizontal displacement xx0from an
initial position x0is given by Eq. 2-15 with a0, which we write as
xx0v0xt.
Because v0xv0cos u0, this becomes
xx0(v0cos u0)t. (4-21)
The Vertical Motion
The vertical motion is the motion we discussed in Module 2-5 for a particle in
free fall. Most important is that the acceleration is constant. Thus, the equations
of Table 2-1 apply, provided we substitute gfor aand switch to ynotation.Then,
for example, Eq. 2-15 becomes
(4-22)
where the initial vertical velocity component v0yis replaced with the equivalent
v0sin u0. Similarly, Eqs. 2-11 and 2-16 become
vyv0sin u0gt (4-23)
and (4-24)vy
2(v0 sin
0)22g(yy0).
(v0 sin
0)t1
2gt 2,
yy0v0yt1
2gt 2
This feature allows us to break up a problem involving two-dimensional motion
into two separate and easier one-dimensional problems, one for the horizontal
motion (with zero acceleration) and one for the vertical motion (with constant
downward acceleration). Here are two experiments that show that the horizontal
motion and the vertical motion are independent.
Two Golf Balls
Figure 4-10 is a stroboscopic photograph of two golf balls, one simply released and
the other shot horizontally by a spring.The golf balls have the same vertical motion,
both falling through the same vertical distance in the same interval of time. The fact
that one ball is moving horizontally while it is falling has no effect on its vertical mo-
tion; that is, the horizontal and vertical motions are independent of each other.
A Great Student Rouser
In Fig. 4-11, a blowgun G using a ball as a projectile is aimed directly at a can sus-
pended from a magnet M. Just as the ball leaves the blowgun, the can is released. If g
(the magnitude of the free-fall acceleration) were zero, the ball would follow the
straight-line path shown in Fig. 4-11 and the can would float in place after the
magnet released it. The ball would certainly hit the can. However, gis not zero,
but the ball still hits the can! As Fig. 4-11 shows, during the time of flight of the
ball, both ball and can fall the same distance hfrom their zero-glocations. The
harder the demonstrator blows, the greater is the ball’s initial speed, the shorter
the flight time, and the smaller the value of h.
73
4-4 PROJECTILE MOTION
As is illustrated in Fig. 4-9 and Eq. 4-23, the vertical velocity component be-
haves just as for a ball thrown vertically upward. It is directed upward initially,
and its magnitude steadily decreases to zero, which marks the maximum height of
the path. The vertical velocity component then reverses direction, and its magni-
tude becomes larger with time.
The Equation of the Path
We can find the equation of the projectile’s path (its trajectory) by eliminating
time tbetween Eqs. 4-21 and 4-22. Solving Eq. 4-21 for tand substituting into
Eq. 4-22, we obtain, after a little rearrangement,
(trajectory). (4-25)
This is the equation of the path shown in Fig. 4-9. In deriving it, for simplicity we
let x00 and y00 in Eqs. 4-21 and 4-22, respectively. Because g,u0, and v0are
constants, Eq. 4-25 is of the form yax bx2, in which aand bare constants.
This is the equation of a parabola, so the path is parabolic.
The Horizontal Range
The horizontal range R of the projectile is the horizontal distance the projectile
has traveled when it returns to its initial height (the height at which it is
launched). To find range R, let us put xx0Rin Eq. 4-21 and yy00 in
Eq. 4-22, obtaining
R(v0cos u0)t
and
Eliminating tbetween these two equations yields
Using the identity sin 2u02 sin u0cos u0(see Appendix E), we obtain
(4-26)
This equation does not give the horizontal distance traveled by a projectile when
the final height is not the launch height. Note that Rin Eq. 4-26 has its maximum
value when sin 2u01, which corresponds to 2u090° or u045°.
Rv0
2
g sin 2
0.
R2v0
2
g sin
0 cos
0.
0(v0 sin
0)t1
2gt 2.
y(tan
0)xgx2
2(v0 cos
0)2
Figure 4-12 The vertical component of this
skateboarder’s velocity is changing but not
the horizontal component, which matches
the skateboard’s velocity. As a result, the
skateboard stays underneath him, allowing
him to land on it.
Jamie Budge
The horizontal range Ris maximum for a launch angle of 45°.
However, when the launch and landing heights differ, as in many sports, a launch
angle of 45° does not yield the maximum horizontal distance.
The Effects of the Air
We have assumed that the air through which the projectile moves has no effect
on its motion. However, in many situations, the disagreement between our calcu-
lations and the actual motion of the projectile can be large because the air resists
(opposes) the motion. Figure 4-13,for example, shows two paths for a fly ball that
leaves the bat at an angle of 60° with the horizontal and an initial speed of
44.7 m/s. Path I (the baseball player’s fly ball) is a calculated path that
approximates normal conditions of play, in air. Path II (the physics professor’s fly
ball) is the path the ball would follow in a vacuum.
Figure 4-13 (I) The path of a fly ball calcu-
lated by taking air resistance into account.
(II) The path the ball would follow in a
vacuum, calculated by the methods of this
chapter. See Table 4-1 for corresponding
data. (Based on “The Trajectory of a Fly
Ball,” by Peter J. Brancazio, The Physics
Teacher, January 1985.)
x
y
60°
v0
I
II
Air reduces
height ... ... and range.
Table 4-1 Two Fly Ballsa
Path I Path II
(Air) (Vacuum)
Range 98.5 m 177 m
Maximum
height 53.0 m 76.8 m
Time
of flight 6.6 s 7.9 s
aSee Fig. 4-13.The launch angle is 60° and the
launch speed is 44.7 m/s.
74 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Additional examples, video, and practice available at WileyPLUS
Sample Problem 4.04 Projectile dropped from airplane
Then Eq. 4-27 gives us
(Answer)
(b) As the capsule reaches the water, what is its velocity ?
KEY IDEAS
(1) The horizontal and vertical components of the capsule’s
velocity are independent. (2) Component vxdoes not change
from its initial value v0xv0cos u0because there is no hori-
zontal acceleration. (3) Component vychanges from its initial
value v0yv0sin u0because there is a vertical acceleration.
Calculations: When the capsule reaches the water,
vxv0cos u0(55.0 m/s)(cos 0°) 55.0 m/s.
Using Eq. 4-23 and the capsule’s time of fall t10.1 s, we
also find that when the capsule reaches the water,
vyv0sin u0gt
(55.0 m/s)(sin 0°) (9.8 m/s2)(10.1 s)
99.0 m/s.
Thus, at the water
(Answer)
From Eq. 3-6, the magnitude and the angle of are
v113 m/s and u60.9°. (Answer)
v
:
v
:(55.0 m/s)i
ˆ(99.0 m/s)j
ˆ.
v
:
tan1555.5 m
500 m 48.0.
In Fig.4-14, a rescue plane flies at 198 km/h (55.0 m/s) and
constant height h500 m toward a point directly over a
victim, where a rescue capsule is to land.
(a) What should be the angle fof the pilot’s line of sight to
the victim when the capsule release is made?
KEY IDEAS
Once released, the capsule is a projectile, so its horizontal
and vertical motions can be considered separately (we need
not consider the actual curved path of the capsule).
Calculations: In Fig.4-14, we see that fis given by
(4-27)
where xis the horizontal coordinate of the victim (and of
the capsule when it hits the water) and h 500 m. We
should be able to find xwith Eq. 4-21:
xx0(v0cos u0)t. (4-28)
Here we know that x00 because the origin is placed at
the point of release. Because the capsule is released and
not shot from the plane, its initial velocity is equal to
the plane’s velocity. Thus, we know also that the initial ve-
locity has magnitude v055.0 m/s and angle u00°
(measured relative to the positive direction of the xaxis).
However, we do not know the time tthe capsule takes to
move from the plane to the victim.
To find t, we next consider the vertical motion and
specifically Eq. 4-22:
(4-29)
Here the vertical displacement yy0of the capsule is
500 m (the negative value indicates that the capsule
moves downward). So,
(4-30)
Solving for t, we find t10.1 s. Using that value in Eq. 4-28
yields
x0(55.0 m/s)(cos 0°)(10.1 s), (4-31)
or x555.5 m.
500 m (55.0 m/s)(sin 0)t1
2(9.8 m/s2)t2.
yy0(v0 sin
0)t1
2gt2.
v
:
0
tan1x
h,
y
θ
φ
O
v0
Trajectory
Line of sight
h
x
v
Figure 4-14 A plane drops a rescue capsule while moving at
constant velocity in level flight. While falling, the capsule
remains under the plane.
Checkpoint 4
A fly ball is hit to the outfield. During its flight (ignore the effects of the air), what
happens to its (a) horizontal and (b) vertical components of velocity? What are the (c)
horizontal and (d) vertical components of its acceleration during ascent, during de-
scent, and at the topmost point of its flight?
75
4-4 PROJECTILE MOTION
Sample Problem 4.05 Launched into the air from a water slide
One of the most dramatic videos on the web (but entirely
fictitious) supposedly shows a man sliding along a long wa-
ter slide and then being launched into the air to land in a
water pool. Let’s attach some reasonable numbers to such
a flight to calculate the velocity with which the man would
have hit the water. Figure 4-15aindicates the launch and
landing sites and includes a superimposed coordinate sys-
tem with its origin conveniently located at the launch site.
From the video we take the horizontal flight distance as
D20.0 m, the flight time as t2.50 s, and the launch an-
gle as 040.0°. Find the magnitude of the velocity at
launch and at landing.
KEY IDEAS
(1) For projectile motion, we can apply the equations for con-
stant acceleration along the horizontal and vertical axes sepa-
rately. (2) Throughout the flight, the vertical acceleration is
ayg9.8 m/s and the horizontal acceleration is .
Calculations: In most projectile problems, the initial chal-
lenge is to figure out where to start. There is nothing wrong
with trying out various equations, to see if we can somehow
get to the velocities. But here is a clue. Because we are going
to apply the constant-acceleration equations separately to
the xand ymotions, we should find the horizontal and verti-
cal components of the velocities at launch and at landing.
For each site, we can then combine the velocity components
to get the velocity.
Because we know the horizontal displacement D
20.0 m, let’s start with the horizontal motion. Since ,ax0
ax0
we know that the horizontal velocity component is con-
stant during the flight and thus is always equal to the hori-
zontal component v0xat launch. We can relate that compo-
nent, the displacement and the flight time t2.50 s
with Eq. 2-15:
(4-32)
Substituting this becomes Eq. 4-21. With
we then write
That is a component of the launch velocity, but we need
the magnitude of the full vector, as shown in Fig. 4-15b,
where the components form the legs of a right triangle and
the full vector forms the hypotenuse. We can then apply a
trig definition to find the magnitude of the full velocity at
launch:
and so
(Answer)
Now let’s go after the magnitude vof the landing veloc-
ity. We already know the horizontal component, which does
not change from its initial value of 8.00 m/s.To find the verti-
cal component vyand because we know the elapsed time t
2.50 s and the vertical acceleration let’s
rewrite Eq. 2-11 as
and then (from Fig.4-15b) as
(4-33)
Substituting ayg,this becomes Eq. 4-23.We can then write
Now that we know both components of the landing velocity,
we use Eq. 3-6 to find the velocity magnitude:
(Answer)19.49 m/s2 19.5 m/s.
2(8.00 m/s)2(17.78 m/s)2
v2vx
2vy
2
17.78 m/s.
vy(10.44 m/s) sin (40.0)(9.8 m/s2)(2.50 s)
vyv0 sin
0ayt.
vyv0yayt
ay9.8 m/s2,
10.44 m/s 10.4 m/s.
v0v0x
cos u0
8.00 m/s
cos 40
cos
0v0x
v0
,
v0x8.00 m/s.
20 m v0x(2.50 s) 1
2 (0)(2.50 s)2
xx0D,ax0,
xx0v0xt1
2axt2.
xx0,
vx
D
θ
0
v0
y
x
Launch Water
pool
(a)
θ
0
v0v0y
v0x
θ
0
v
vy
v0x
(b)(c)
Landing
velocity
Launch
velocity
Figure 4-15 (a) Launch from a water slide, to land in a water pool.
The velocity at (b) launch and (c) landing.
Additional examples, video, and practice available at WileyPLUS
76 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Uniform Circular Motion
A particle is in uniform circular motion if it travels around a circle or a circular
arc at constant (uniform) speed. Although the speed does not vary, the particle is
accelerating because the velocity changes in direction.
Figure 4-16 shows the relationship between the velocity and acceleration
vectors at various stages during uniform circular motion. Both vectors have
constant magnitude, but their directions change continuously. The velocity is
always directed tangent to the circle in the direction of motion. The accelera-
tion is always directed radially inward. Because of this, the acceleration associ-
ated with uniform circular motion is called a centripetal (meaning “center seek-
ing”) acceleration. As we prove next, the magnitude of this acceleration is
(centripetal acceleration), (4-34)
where ris the radius of the circle and vis the speed of the particle.
In addition, during this acceleration at constant speed, the particle travels the
circumference of the circle (a distance of 2pr) in time
(period). (4-35)
Tis called the period of revolution, or simply the period, of the motion. It is, in
general, the time for a particle to go around a closed path exactly once.
Proof of Eq. 4-34
To find the magnitude and direction of the acceleration for uniform circular
motion, we consider Fig. 4-17. In Fig. 4-17a, particle pmoves at constant speed
varound a circle of radius r. At the instant shown, phas coordinates xpand yp.
Recall from Module 4-2 that the velocity of a moving particle is always
tangent to the particle’s path at the particle’s position. In Fig. 4-17a, that means
is perpendicular to a radius rdrawn to the particle’s position.Then the angle
uthat makes with a vertical at pequals the angle uthat radius rmakes with
the xaxis.
v
:
v
:
v
:
T2
r
v
av2
r
a
:
Figure 4-16 Velocity and acceleration
vectors for uniform circular motion.
v
v
v
a
a
a
The acceleration vector
always points toward the
center.
The velocity
vector is always
tangent to the path.
4-5 UNIFORM CIRCULAR MOTION
4.17 Apply the relationships between the radius of the circu-
lar path, the period, the particle’s speed, and the particle’s
acceleration magnitude.
Learning Objectives
After reading this module, you should be able to . . .
4.16 Sketch the path taken in uniform circular motion and ex-
plain the velocity and acceleration vectors (magnitude and
direction) during the motion.
arc, and is said to be centripetal. The time for the particle to
complete a circle is
Tis called the period of revolution, or simply the period, of the
motion.
T2
r
v
.
a
:
Key Ideas
●If a particle travels along a circle or circular arc of radius rat
constant speed v, it is said to be in uniform circular motion
and has an acceleration of constant magnitude
The direction of is toward the center of the circle or circulara
:
av2
r.
a
:
77
4-5 UNIFORM CIRCULAR MOTION
Figure 4-17 Particle pmoves in counter-
clockwise uniform circular motion. (a) Its
position and velocity at a certain
instant. (b) Velocity . (c) Acceleration .a
:
v
:
v
:
y
x
θ
θ
p
yp
r
xp
v
(a)
y
x
θ
vx
vy
v
(b)
y
x
φ
ax
ay
a
(c)
The scalar components of are shown in Fig. 4-17b.With them, we can write
the velocity as
. (4-36)
Now, using the right triangle in Fig. 4-17a, we can replace sin uwith yp/rand
cos uwith xp/rto write
(4-37)
To find the acceleration of particle p, we must take the time derivative of this
equation.Noting that speed vand radius rdo not change with time,we obtain
(4-38)
Now note that the rate dyp/dt at which ypchanges is equal to the velocity
component vy. Similarly, dxp/dt vx, and, again from Fig. 4-17b, we see that vx
vsin uand vyvcos u. Making these substitutions in Eq. 4-38, we find
. (4-39)
This vector and its components are shown in Fig. 4-17c. Following Eq. 3-6, we find
as we wanted to prove.To orient , we find the angle fshown in Fig. 4-17c:
.
Thus, fu, which means that is directed along the radius rof Fig. 4-17a,
toward the circle’s center, as we wanted to prove.
a
:
tan
ay
ax
(v2/r) sin
(v2/r) cos
tan
a
:
a2ax
2ay
2v2
r2(cos
)2(sin
)2v2
r11v2
r,
a
:
v2
r cos
i
ˆ
v2
r sin
j
ˆ
a
:dv
:
dt
v
r
dyp
dt
i
ˆ
v
r
dxp
dt
j
ˆ.
a
:
v
:
vyp
r
i
ˆ
vxp
r
j
ˆ.
v
:vxi
ˆvyj
ˆ(v sin
)i
ˆ(v cos
)j
ˆ
v
:
v
:
Checkpoint 5
An object moves at constant speed along a circular path in a horizontal xy plane, with
the center at the origin.When the object is at x2 m, its velocity is (4 m/s) . Give
the object’s (a) velocity and (b) acceleration at y2m.
j
ˆ
Sample Problem 4.06 Top gun pilots in turns
KEY IDEAS
We assume the turn is made with uniform circular motion.
Then the pilot’s acceleration is centripetal and has magni-
tude agiven by Eq. 4-34 (av2/R), where Ris the circle’s
radius.Also, the time required to complete a full circle is the
period given by Eq. 4-35 (T2pR/v).
Calculations: Because we do not know radius R, let’s solve
Eq. 4-35 for Rand substitute into Eq. 4-34.We find
To get the constant speed v, let’s substitute the components
of the initial velocity into Eq. 3-6:
v2(400 m/s)2(500 m/s)2 640.31 m/s.
a2
v
T.
“Top gun” pilots have long worried about taking a turn too
tightly. As a pilot’s body undergoes centripetal acceleration,
with the head toward the center of curvature, the blood pres-
sure in the brain decreases, leading to loss of brain function.
There are several warning signs. When the centripetal
acceleration is 2gor 3g, the pilot feels heavy. At about 4g,
the pilot’s vision switches to black and white and narrows to
“tunnel vision.” If that acceleration is sustained or in-
creased, vision ceases and, soon after, the pilot is uncon-
scious—a condition known as g-LOC for “g-induced loss of
consciousness.”
What is the magnitude of the acceleration, in gunits, of
a pilot whose aircraft enters a horizontal circular turn with a
velocity of (400ˆ
i 500ˆ
j) m/s and 24.0 s later leaves the
turn with a velocity of ( 400ˆ
i 500ˆ
j) m/s?v
:
f
v
:
i
78 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
4-6 RELATIVE MOTION IN ONE DIMENSION
frames that move relative to each other at constant velocity
and along a single axis.
Learning Objective
After reading this module, you should be able to . . .
4.18 Apply the relationship between a particle’s position, ve-
locity, and acceleration as measured from two reference
where is the velocity of Bwith respect to A. Both ob-
servers measure the same acceleration for the particle:
a
:
PA a
:
PB.
v
:
BA
v
:
PA v
:
PB v
:
BA,
Key Idea
●When two frames of reference Aand Bare moving relative
to each other at constant velocity, the velocity of a particle P
as measured by an observer in frame Ausually differs from
that measured from frame B. The two measured velocities are
related by
Relative Motion in One Dimension
Suppose you see a duck flying north at 30 km/h.To another duck flying alongside,
the first duck seems to be stationary. In other words, the velocity of a particle de-
pends on the reference frame of whoever is observing or measuring the velocity.
For our purposes, a reference frame is the physical object to which we attach our
coordinate system. In everyday life, that object is the ground. For example, the
speed listed on a speeding ticket is always measured relative to the ground. The
speed relative to the police officer would be different if the officer were moving
while making the speed measurement.
Suppose that Alex (at the origin of frame Ain Fig. 4-18) is parked by the side
of a highway, watching car P(the “particle”) speed past. Barbara (at the origin of
frame B) is driving along the highway at constant speed and is also watching car P.
Suppose that they both measure the position of the car at a given moment. From
Fig.4-18 we see that
xPA xPB xBA. (4-40)
The equation is read: “The coordinate xPA of Pas measured by A is equal to the
coordinate xPB of Pas measured by B plus the coordinate xBA of Bas measured
by A.” Note how this reading is supported by the sequence of the subscripts.
Taking the time derivative of Eq. 4-40, we obtain
Thus, the velocity components are related by
vPA vPB vBA. (4-41)
This equation is read: “The velocity vPA of Pas measured by A is equal to the
d
dt (xPA)d
dt (xPB)d
dt (xBA).
Figure 4-18 Alex (frame A) and Barbara
(frame B) watch car P, as both Band P
move at different velocities along the com-
mon xaxis of the two frames. At the
instant shown, xBA is the coordinate of B
in the Aframe. Also, Pis at coordinate xPB
in the Bframe and coordinate xPA xPB
xBA in the Aframe.
x
Frame AFrame B
vBA
P
x
y
y
xPA =x
PB +x
BA
xBA
xPB
Frame B moves past
frame A while both
observe P.
To find the period Tof the motion, first note that the final
velocity is the reverse of the initial velocity. This means the
aircraft leaves on the opposite side of the circle from the ini-
tial point and must have completed half a circle in the given
24.0 s. Thus a full circle would have taken T48.0 s.
Substituting these values into our equation for a, we find
(Answer)a2
(640.31 m/s)
48.0 s 83.81 m/s28.6g.
Additional examples, video, and practice available at WileyPLUS
79
4-6 RELATIVE MOTION IN ONE DIMENSION
velocity vPB of Pas measured by B plus the velocity vBA of Bas measured by A.”
The term vBA is the velocity of frame Brelative to frame A.
Here we consider only frames that move at constant velocity relative to
each other. In our example, this means that Barbara (frame B) drives always at
constant velocity vBA relative to Alex (frame A). Car P(the moving particle),
however, can change speed and direction (that is, it can accelerate).
To relate an acceleration of Pas measured by Barbara and by Alex, we take
the time derivative of Eq. 4-41:
Because vBA is constant, the last term is zero and we have
aPA aPB. (4-42)
In other words,
d
dt (vPA)d
dt (vPB)d
dt (vBA).
Observers on different frames of reference that move at constant velocity relative
to each other will measure the same acceleration for a moving particle.
Sample Problem 4.07 Relative motion, one dimensional, Alex and Barbara
to relate the acceleration to the initial and final velocities
of P.
Calculation: The initial velocity of Prelative to Alex is
vPA 78 km/h and the final velocity is 0.Thus, the acceler-
ation relative to Alex is
(Answer)
(c) What is the acceleration aPB of car Prelative to Barbara
during the braking?
KEY IDEA
To calculate the acceleration of car P relative to Barbara, we
must use the car’s velocities relative to Barbara.
Calculation: We know the initial velocity of Prelative to
Barbara from part (a) (vPB 130 km/h).The final veloc-
ity of Prelative to Barbara is 52 km/h (because this is
the velocity of the stopped car relative to the moving
Barbara).Thus,
(Answer)
Comment: We should have foreseen this result: Because
Alex and Barbara have a constant relative velocity, they
must measure the same acceleration for the car.
2.2 m/s2.
aPB vv0
t52 km/h (130 km/h)
10 s
1 m/s
3.6 km/h
2.2 m/s2.
aPA vv0
t0(78 km/h)
10 s
1 m/s
3.6 km/h
In Fig. 4-18, suppose that Barbara’s velocity relative to Alex
is a constant vBA 52 km/h and car Pis moving in the nega-
tive direction of the xaxis.
(a) If Alex measures a constant vPA 78 km/h for car P,
what velocity vPB will Barbara measure?
KEY IDEAS
We can attach a frame of reference Ato Alex and a frame of
reference Bto Barbara. Because the frames move at constant
velocity relative to each other along one axis, we can use
Eq. 4-41 (vPA vPB vBA) to relate vPB to vPA and vBA .
Calculation: We find
78 km/h vPB 52 km/h.
Thus, vPB 130 km/h. (Answer)
Comment: If car Pwere connected to Barbara’s car by a
cord wound on a spool, the cord would be unwinding at
a speed of 130 km/h as the two cars separated.
(b) If car Pbrakes to a stop relative to Alex (and thus rela-
tive to the ground) in time t10 s at constant acceleration,
what is its acceleration aPA relative to Alex?
KEY IDEAS
To calculate the acceleration of car P relative to Alex, we
must use the car’s velocities relative to Alex. Because the
acceleration is constant, we can use Eq. 2-11 (vv0at)
Additional examples, video, and practice available at WileyPLUS
80 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
4-7 RELATIVE MOTION IN TWO DIMENSIONS
frames that move relative to each other at constant velocity
and in two dimensions.
Learning Objective
After reading this module, you should be able to . . .
4.19 Apply the relationship between a particle’s position, ve-
locity, and acceleration as measured from two reference
where is the velocity of Bwith respect to A. Both
observers measure the same acceleration for the particle:
a
:
PA a
:
PB.
v
:
BA
v
:
PA v
:
PB v
:
BA,
Key Idea
●When two frames of reference Aand Bare moving relative
to each other at constant velocity, the velocity of a particle
Pas measured by an observer in frame Ausually differs from
that measured from frame B. The two measured velocities are
related by
Relative Motion in Two Dimensions
Our two observers are again watching a moving particle Pfrom the origins of refer-
ence frames Aand B, while Bmoves at a constant velocity relative to A.(The
corresponding axes of these two frames remain parallel.) Figure 4-19 shows a cer-
tain instant during the motion.At that instant, the position vector of the origin of B
relative to the origin of Ais .Also, the position vectors of particle Pare rela-
tive to the origin of Aand relative to the origin of B. From the arrangement of
heads and tails of those three position vectors, we can relate the vectors with
(4-43)
By taking the time derivative of this equation, we can relate the velocities
and of particle Prelative to our observers:
(4-44)
By taking the time derivative of this relation, we can relate the accelerations
and of the particle Prelative to our observers. However, note that because
is constant, its time derivative is zero.Thus, we get
(4-45)
As for one-dimensional motion, we have the following rule: Observers on differ-
ent frames of reference that move at constant velocity relative to each other will
measure the same acceleration for a moving particle.
a
:
PA a
:
PB.
v
:
BA
a
:
PB
a
:
PA
v
:
PA v
:
PB v
:
BA.
v
:
PB
v
:
PA
r
:
PA r
:
PB r
:
BA.
r
:
PB
r
:
PA
r
:
BA
v
:
BA
Figure 4-19 Frame Bhas the constant
two-dimensional velocity relative to
frame A. The position vector of Brelative
to Ais . The position vectors of parti-
cle Pare relative to Aand
relative to B.
r
:
PB
r
:
PA
r
:
BA
v
:
BA
x
x
y
y
rPB
rPA
rBA
Frame B
Frame A
vBA
P
Sample Problem 4.08 Relative motion, two dimensional, airplanes
In Fig. 4-20a, a plane moves due east while the pilot points
the plane somewhat south of east, toward a steady wind that
blows to the northeast. The plane has velocity relative
to the wind, with an airspeed (speed relative to the wind)
of 215 km/h, directed at angle usouth of east. The wind
has velocity relative to the ground with speed
65.0 km/h, directed 20.0° east of north. What is the magni-
tude of the velocity of the plane relative to the ground,
and what is ?
v
:
PG
v
:
WG
v
:
PW
KEY IDEAS
The situation is like the one in Fig. 4-19. Here the moving par-
ticle Pis the plane, frame Ais attached to the ground (call it
G), and frame Bis “attached” to the wind (call it W).We need
a vector diagram like Fig. 4-19 but with three velocity vectors.
Calculations: First we construct a sentence that relates the
three vectors shown in Fig.4-20b:
81
REVIEW & SUMMARY
velocity of plane velocity of plane velocity of wind
relative to ground relative to wind relative to ground.
(PG)(PW)(WG)
This relation is written in vector notation as
(4-46)
We need to resolve the vectors into components on the co-
ordinate system of Fig. 4-20band then solve Eq. 4-46 axis by
axis. For the ycomponents, we find
vPG,y vPW,y vWG,y
or 0 (215 km/h) sin u(65.0 km/h)(cos 20.0°).
Solving for ugives us
(Answer)
Similarly, for the xcomponents we find
vPG,x vPW,x vWG,x.
Here, because is parallel to the xaxis, the component
vPG,x is equal to the magnitude vPG. Substituting this nota-
tion and the value u16.5°, we find
vPG (215 km/h)(cos 16.5°) (65.0 km/h)(sin 20.0°)
228 km/h. (Answer)
v
:
PG
sin1(65.0 km/h)(cos 20.0)
215 km/h 16.5.
v
:
PG v
:
PW v
:
WG.
θ
θ
vPG
vPW vWG
vPG
vPW vWG
N
y
N
E
20°
x
(a)
(b)
This is the plane's actual
direction of travel.
This is the wind
direction.
The actual direction
is the vector sum of
the other two vectors
(head-to-tail arrangement).
This is the plane's
orientation.
Figure 4-20 A plane flying in a wind.
Additional examples, video, and practice available at WileyPLUS
Review & Summary
Position Vector The location of a particle relative to the ori-
gin of a coordinate system is given by a position vector , which in
unit-vector notation is
(4-1)
Here x,y, and zare the vector components of position vector ,
and x,y, and zare its scalar components (as well as the coordinates
of the particle). A position vector is described either by a magni-
tude and one or two angles for orientation, or by its vector or
scalar components.
Displacement If a particle moves so that its position vector
changes from to , the particle’s displacement is
(4-2)
The displacement can also be written as
(4-3)
xyz. (4-4)
Average Velocity and Instantaneous Velocity If a parti-
cle undergoes a displacement in time interval t, its average ve-
locity for that time interval is
(4-8)v
:
avg r
:
t.
v
:
avg
r
:
k
ˆ
j
ˆ
i
ˆ
r
:(x2x1)i
ˆ(y2y1)j
ˆ(z2z1)k
ˆ
r
:r
:
2r
:
1.
r
:
r
:
2
r
:
1
r
:
k
ˆ
j
ˆ
i
ˆ
r
:xi
ˆyj
ˆzk
ˆ.
r
:As tin Eq. 4-8 is shrunk to 0, reaches a limit called either the
velocity or the instantaneous velocity :
(4-10)
which can be rewritten in unit-vector notation as
(4-11)
where vxdx/dt, vydy/dt, and vzdz /dt. The instantaneous
velocity of a particle is always directed along the tangent to the
particle’s path at the particle’s position.
Average Acceleration and Instantaneous Acceleration
If a particle’s velocity changes from to in time interval t,its
average acceleration during tis
(4-15)
As tin Eq. 4-15 is shrunk to 0, reaches a limiting value calleda
:
avg
a
:
avg v
:
2v
:
1
tv
:
t.
v
:
2
v
:
1
v
:
v
:vxi
ˆvyj
ˆvzk
ˆ,
v
:dr
:
dt ,
v
:
v
:
avg
either the acceleration or the instantaneous acceleration :
(4-16)
In unit-vector notation,
(4-17)
where axdvx/dt, aydvy/dt, and azdvz/dt.
a
:axi
ˆayj
ˆazk
ˆ,
a
:dv
:
dt .
a
:
82 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Figure 4-23 Question 5.
(a)(b)(c)
Projectile Motion Projectile motion is the motion of a particle
that is launched with an initial velocity . During its flight, the par-
ticle’s horizontal acceleration is zero and its vertical acceleration is
the free-fall acceleration g. (Upward is taken to be a positive di-
rection.) If is expressed as a magnitude (the speed v0) and an an-
gle u0(measured from the horizontal), the particle’s equations of
motion along the horizontal xaxis and vertical yaxis are
v
:
0
v
:
0
Questions
1Figure 4-21 shows the path taken by
a skunk foraging for trash food, from
initial point i. The skunk took the same
time Tto go from each labeled point to
the next along its path. Rank points a,b,
and caccording to the magnitude of the
average velocity of the skunk to reach
them from initial point i, greatest first.
2Figure 4-22 shows the initial posi-
tion iand the final position fof a parti-
cle. What are the (a) initial position
vector and (b) final position vector , both in unit-vector nota-
rf
:
r
:
i
Figure 4-21
Question 1.
aib c
Figure 4-22 Question 2.
z
x
i
f
y
4 m 4 m
1 m
2 m
3 m
3 m
3 m 5 m
twice as long as at 45º. Does that result mean that the air density at
high altitudes increases with altitude or decreases?
4You are to launch a rocket, from just above the ground, with
one of the following initial velocity vectors: (1) ,
(2) , (3) , (4) . In
your coordinate system, xruns along level ground and yincreases
upward. (a) Rank the vectors according to the launch speed of the
projectile, greatest first. (b) Rank the vectors according to the time
of flight of the projectile, greatest first.
5Figure 4-23 shows three situations in which identical projectiles
are launched (at the same level) at identical initial speeds and an-
gles. The projectiles do not land on the same terrain, however.
Rank the situations according to the final speeds of the projectiles
just before they land, greatest first.
20i
ˆ70j
ˆ
v
:
0v
:
020i
ˆ70j
ˆ
v
:
020i
ˆ70j
ˆ20i
ˆ70j
ˆ
v
:
0
Uniform Circular Motion If a particle travels along a circle or
circular arc of radius rat constant speed v, it is said to be in uniform
circular motion and has an acceleration of constant magnitude
(4-34)
The direction of is toward the center of the circle or circular arc,
and is said to be centripetal. The time for the particle to complete
a circle is
. (4-35)
Tis called the period of revolution, or simply the period, of the
motion.
Relative Motion When two frames of reference Aand Bare
moving relative to each other at constant velocity, the velocity of a par-
ticle Pas measured by an observer in frame Ausually differs from that
measured from frame B.The two measured velocities are related by
(4-44)
where is the velocity of Bwith respect to A. Both observers
measure the same acceleration for the particle:
(4-45)a
:
PA a
:
PB.
v
:
BA
v
:
PA v
:
PB v
:
BA,
T2
r
v
a
:
a
:
av2
r.
a
:
xx0(v0cos u0)t, (4-21)
, (4-22)
vyv0sin u0gt, (4-23)
. (4-24)
The trajectory (path) of a particle in projectile motion is parabolic
and is given by
, (4-25)
if x0and y0of Eqs. 4-21 to 4-24 are zero. The particle’s horizontal
range R, which is the horizontal distance from the launch point to
the point at which the particle returns to the launch height, is
(4-26)Rv2
0
g sin 2
0.
y(tan
0)xgx2
2(v0 cos
0)2
v2
y(v0 sin
0)22g(yy0)
yy0(v0 sin
0)t1
2gt2
3When Paris was shelled from 100 km away with the WWI
long-range artillery piece “Big Bertha,” the shells were fired at an
angle greater than 45º to give them a greater range, possibly even
6The only good use of a fruitcake
is in catapult practice. Curve 1 in
Fig. 4-24 gives the height yof a cata-
pulted fruitcake versus the angle u
between its velocity vector and its
acceleration vector during flight. (a)
Which of the lettered points on that
curve corresponds to the landing of
the fruitcake on the ground? (b)
Curve 2 is a similar plot for the same
y
θ
AB
2
1
Figure 4-24 Question 6.
tion? (c) What is the xcomponent of displacement ?r
:
83
QUESTIONS
launch speed but for a different launch angle. Does the fruitcake
now land farther away or closer to the launch point?
7An airplane flying horizontally at a constant speed of 350 km/h
over level ground releases a bundle of food supplies. Ignore the ef-
fect of the air on the bundle.What are the bundle’s initial (a) verti-
cal and (b) horizontal components of velocity? (c) What is its hori-
zontal component of velocity just before hitting the ground? (d) If
the airplane’s speed were, instead, 450 km/h, would the time of fall
be longer,shorter,or the same?
8In Fig. 4-25, a cream tangerine is thrown up past windows 1, 2,
and 3, which are identical in size and regularly spaced vertically.
Rank those three windows according to (a) the time the cream tan-
gerine takes to pass them and (b) the average speed of the cream
tangerine during the passage, greatest first.
The cream tangerine then moves down past windows 4, 5,
and 6, which are identical in size and irregularly spaced horizon-
tally. Rank those three windows according to (c) the time the
cream tangerine takes to pass them and (d) the average speed of
the cream tangerine during the passage, greatest first.
11 Figure 4-28 shows four tracks (either half- or quarter-circles)
that can be taken by a train, which moves at a constant speed.
Rank the tracks according to the magnitude of a train’s accelera-
tion on the curved portion, greatest first.
1
2
34
5
6
Figure 4-25 Question 8.
R
θ
0
b
a
c
Figure 4-27 Question 10.
3
4
2
1
Figure 4-28 Question 11.
x
y
θ
r
P
Figure 4-29 Question 12.
12 3
Figure 4-26 Question 9.
10 A ball is shot from ground level over level ground at a certain
initial speed. Figure 4-27 gives the range Rof the ball versus its
launch angle u0. Rank the three lettered points on the plot accord-
ing to (a) the total flight time of the ball and (b) the ball’s speed at
maximum height, greatest first.
12 In Fig. 4-29, particle Pis in uniform circular motion, cen-
tered on the origin of an xy coordinate system. (a) At what values
of uis the vertical component ryof the position vector greatest in
magnitude? (b) At what values of uis the vertical component vy
of the particle’s velocity greatest in magnitude? (c) At what val-
ues of uis the vertical component ayof the particle’s acceleration
greatest in magnitude?
13 (a) Is it possible to be accelerating while traveling at constant
speed? Is it possible to round a curve with (b) zero acceleration and
(c) a constant magnitude of acceleration?
14 While riding in a moving car, you toss an egg directly upward.
Does the egg tend to land behind you, in front of you, or back in your
hands if the car is (a) traveling at a constant speed, (b) increasing in
speed,and (c) decreasing in speed?
15 A snowball is thrown from ground level (by someone in a
hole) with initial speed v0at an angle of 45° relative to the (level)
ground, on which the snowball later lands. If the launch angle is in-
creased, do (a) the range and (b) the flight time increase, decrease,
or stay the same?
16 You are driving directly behind a pickup truck, going at the
same speed as the truck. A crate falls from the bed of the truck to
the road. (a) Will your car hit the crate before the crate hits the
road if you neither brake nor swerve? (b) During the fall, is the
horizontal speed of the crate more than, less than, or the same as
that of the truck?
17 At what point in the path of a projectile is the speed a minimum?
18 In shot put, the shot is put (thrown) from above the athlete’s
shoulder level. Is the launch angle that produces the greatest range
45°, less than 45°, or greater than 45°?
9Figure 4-26 shows three paths for a football kicked from ground
level. Ignoring the effects of air, rank the paths according to (a) time
of flight, (b) initial vertical velocity component, (c) initial horizontal
velocity component, and (d) initial speed, greatest first.
84 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
Module 4-1 Position and Displacement
•1 The position vector for an electron is
. (a) Find the magnitude of . (b) Sketch the
vector on a right-handed coordinate system.
•2 A watermelon seed has the following coordinates: x5.0 m,
y8.0 m, and z0 m. Find its position vector (a) in unit-vector no-
tation and as (b) a magnitude and (c) an angle relative to the positive
direction of the xaxis. (d) Sketch the vector on a right-handed coor-
dinate system. If the seed is moved to the xyz coordinates (3.00 m,
0 m, 0 m), what is its displacement (e) in unit-vector notation and as
(f) a magnitude and (g) an angle relative to the positive xdirection?
•3 A positron undergoes a displacement ,
ending with the position vector , in meters. What
was the positron’s initial position vector?
••4 The minute hand of a wall clock measures 10 cm from its tip to
the axis about which it rotates. The magnitude and angle of the dis-
placement vector of the tip are to be determined for three time inter-
vals. What are the (a) magnitude and (b) angle from a quarter after
the hour to half past, the (c) magnitude and (d) angle for the next half
hour,and the (e) magnitude and (f) angle for the hour after that?
Module 4-2 Average Velocity and Instantaneous Velocity
•5 A train at a constant 60.0 km/h moves east for 40.0 min,
then in a direction 50.0° east of due north for 20.0 min, and then
west for 50.0 min. What are the (a) magnitude and (b) angle of its
average velocity during this trip?
•6 An electron’s position is given by ,
with tin seconds and in meters. (a) In unit-vector notation, what
is the electron’s velocity ? At t2.00 s, what is (b) in unit-
vector notation and as (c) a magnitude and (d) an angle relative to
the positive direction of the xaxis?
•7 An ion’s position vector is initially ,
and 10 s later it is , all in meters. In unit-
vector notation, what is its during the 10 s?
••8 A plane flies 483 km east from city Ato city Bin 45.0 min and
then 966 km south from city Bto city Cin 1.50 h. For the total trip,
what are the (a) magnitude and (b) direction of the plane’s dis-
placement, the (c) magnitude
and (d) direction of its aver-
age velocity, and (e) its aver-
age speed?
••9 Figure 4-30 gives the
path of a squirrel moving
about on level ground, from
point A(at time t0), to
points B(at t5.00 min), C
(at t10.0 min), and finally D
(at t15.0 min). Consider the
average velocities of the squir-
rel from point Ato each of the
other three points. Of them,
what are the (a) magnitude
v
:
avg
r
:2.0i
ˆ8.0j
ˆ2.0k
ˆ2.0k
ˆ
6.0j
ˆr
:5.0i
ˆ
v
:
v
:(t)
r
:
4.00t2j
ˆ2.00k
ˆ
r
:3.00ti
ˆ
SSM
r
:3.0j
ˆ4.0k
ˆ3.0j
ˆ6.0k
ˆ
r
:2.0i
ˆ
r
:
(3.0 m)j
ˆ(2.0 m)k
ˆr
:(5.0 m)i
ˆ
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
and (b) angle of the one with the
least magnitude and the (c) magni-
tude and (d) angle of the one with
the greatest magnitude?
•••10 The position vector
locates a
particle as a function of time t.
Vector is in meters, tis in seconds,
and factors eand fare constants.
Figure 4-31 gives the angle uof the
particle’s direction of travel as a
function of t(uis measured from
the positive xdirection). What are (a) eand (b) f, including units?
Module 4-3 Average Acceleration and
Instantaneous Acceleration
•11 The position of a particle moving in an xy plane is givenr
:
r
:
r
:5.00ti
ˆ(et ft2)j
ˆ
D
CA
B
25 50
50
25
0
–25
–50
y (m)
x (m)
Figure 4-30 Problem 9.
θ
20°
0°
–20°
10 20
t (s)
Figure 4-31 Problem 10.
by , with in meters and t
in seconds. In unit-vector notation, calculate (a) , (b) , and (c)
for t2.00 s. (d) What is the angle between the positive direction
of the xaxis and a line tangent to the particle’s path at t2.00 s?
•12 At one instant a bicyclist is 40.0 m due east of a park’s flag-
pole, going due south with a speed of 10.0 m/s.Then 30.0 s later, the
cyclist is 40.0 m due north of the flagpole, going due east with a
speed of 10.0 m/s. For the cyclist in this 30.0 s interval, what are the
(a) magnitude and (b) direction of the displacement, the (c) magni-
tude and (d) direction of the average velocity, and the (e) magni-
tude and (f) direction of the average acceleration?
•13 A particle moves so that its position (in meters) as
SSM
a
:
v
:
r
:
r
:
r
:(2.00t35.00t)i
ˆ(6.00 7.00t4)j
ˆ
given by , with in meters per second
and t(> 0) in seconds. (a) What is the acceleration when t3.0 s?
(b) When (if ever) is the acceleration zero? (c) When (if ever) is
the velocity zero? (d) When (if ever) does the speed equal
10 m/s?
••17 A cart is propelled over an xy plane with acceleration compo-
nents ax4.0 m/s2and ay2.0 m/s2. Its initial velocity has com-
ponents v0x8.0 m/s and v0y12 m/s. In unit-vector notation, what
is the velocity of the cart when it reaches its greatest ycoordinate?
••18 A moderate wind accelerates a pebble over a horizontal xy
plane with a constant acceleration .(7.00 m/s2)j
ˆ
a
:(5.00 m/s2)i
ˆ
v
:
v
:(6.0t4.0t2)i
ˆ8.0j
ˆ
a function of time (in seconds) is . Write expres-
sions for (a) its velocity and (b) its acceleration as functions of time.
•14 A proton initially has and then
4.0 s later has (in meters per second). For
that 4.0 s, what are (a) the proton’s average acceleration in unit-
vector notation, (b) the magnitude of , and (c) the angle between
and the positive direction of the xaxis?
••15 A particle leaves the origin with an initial veloc-
ity and a constant acceleration
. When it reaches its maximum xcoordinate, what are
its (a) velocity and (b) position vector?
••16 The velocity of a particle moving in the xy plane isv
:
0.500j
ˆ) m/s2
a
:(1.00i
ˆ(3.00i
ˆ) m/sv
:
ILWSSM
a
:
avg
a
:
avg
a
:
avg
v
:2.0i
ˆ2.0j
ˆ5.0k
ˆ
v
:4.0i
ˆ2.0j
ˆ3.0k
ˆ
r
:i
ˆ4t2j
ˆtk
ˆ
85
PROBLEMS
••27 A certain airplane has a
speed of 290.0 km/h and is diving
at an angle of 30.0° below the
horizontal when the pilot releases
a radar decoy (Fig. 4-33). The hori-
zontal distance between the re-
lease point and the point where
the decoy strikes the ground is d
700 m. (a) How long is the decoy in
the air? (b) How high was the re-
lease point?
••28 In Fig. 4-34, a stone is pro-
jected at a cliff of height hwith an initial speed of 42.0 m/s directed
at angle u060.0° above the horizontal. The stone strikes at A,
5.50 s after launching. Find (a) the height hof the cliff, (b) the
speed of the stone just before impact at A, and (c) the maximum
height Hreached above the ground.
ILW
At time t0, the velocity is (4.00 m/s)i. What are the (a) magni-
tude and (b) angle of its velocity when it has been displaced by
12.0 m parallel to the xaxis?
•••19 The acceleration of a particle moving only on a horizontal
xy plane is given by , where is in meters per second-
squared and tis in seconds. At t0, the position vector
locates the particle, which then has the
velocity vector . At t4.00 s, what
are (a) its position vector in unit-vector notation and (b) the angle
between its direction of travel and the positive direction of the
xaxis?
•••20 In Fig. 4-32, particle A
moves along the line y30 m
with a constant velocity of mag-
nitude 3.0 m/s and parallel to the
xaxis. At the instant particle A
passes the yaxis, particle Bleaves
the origin with a zero initial speed
and a constant acceleration of
magnitude 0.40 m/s2. What angle u
between and the positive direc-
tion of the yaxis would result in a
collision?
Module 4-4 Projectile Motion
•21 A dart is thrown horizontally with an initial speed of
10 m/s toward point P, the bull’s-eye on a dart board. It hits at
point Qon the rim, vertically below P, 0.19 s later. (a) What is the
distance PQ? (b) How far away from the dart board is the dart
released?
•22 A small ball rolls horizontally off the edge of a tabletop that
is 1.20 m high. It strikes the floor at a point 1.52 m horizontally
from the table edge. (a) How long is the ball in the air? (b) What is
its speed at the instant it leaves the table?
•23 A projectile is fired horizontally from a gun that is
45.0 m above flat ground, emerging from the gun with a speed of
250 m/s. (a) How long does the projectile remain in the air? (b) At
what horizontal distance from the firing point does it strike the
ground? (c) What is the magnitude of the vertical component of its
velocity as it strikes the ground?
•24 In the 1991 World Track and Field Championships in
Tokyo, Mike Powell jumped 8.95 m, breaking by a full 5 cm the
23-year long-jump record set by Bob Beamon. Assume that
Powell’s speed on takeoff was 9.5 m/s (about equal to that of a
sprinter) and that g9.80 m/s2in Tokyo. How much less was
Powell’s range than the maximum possible range for a particle
launched at the same speed?
•25 The current world-record motorcycle jump is 77.0 m,
set by Jason Renie. Assume that he left the take-off ramp at
12.0º to the horizontal and that the take-off and landing
heights are the same. Neglecting air drag, determine his take-off
speed.
•26 A stone is catapulted at time t0, with an initial velocity of
magnitude 20.0 m/s and at an angle of 40.0° above the horizontal.
What are the magnitudes of the (a) horizontal and (b) vertical
components of its displacement from the catapult site at t1.10 s?
Repeat for the (c) horizontal and (d) vertical components at
t1.80 s, and for the (e) horizontal and (f) vertical components at
t5.00 s.
a
:
a
:
v
:
v
:(5.00 m/s)i
ˆ(2.00 m/s)j
ˆ
r
:(20.0 m)i
ˆ(40.0 m)j
ˆ
a
:
a
:3ti
ˆ4tj
ˆ
i
ˆ
x
B
A
y
θ
v
a
Figure 4-32 Problem 20.
θ
d
Figure 4-33 Problem 27.
0
H
h
A
θ
Figure 4-34 Problem 28.
••29 A projectile’s launch speed is five times its speed at maxi-
mum height. Find launch angle .
••30 A soccer ball is kicked from the ground with an initial
speed of 19.5 m/s at an upward angle of 45°.A player 55 m away in
the direction of the kick starts running to meet the ball at that in-
stant. What must be his average speed if he is to meet the ball just
before it hits the ground?
••31 In a jump spike, a volleyball player slams the ball from
overhead and toward the opposite floor. Controlling the angle of
the spike is difficult. Suppose a ball is spiked from a height of 2.30
m with an initial speed of 20.0 m/s at a downward angle of 18.00°.
How much farther on the opposite floor would it have landed if the
downward angle were, instead, 8.00°?
••32 You throw a ball toward a
wall at speed 25.0 m/s and at angle
40.0° above the horizontal
(Fig. 4-35). The wall is distance d
22.0 m from the release point of the
ball. (a) How far above the release
point does the ball hit the wall?
What are the (b) horizontal and
(c) vertical components of its velocity as it hits the wall? (d) When
it hits, has it passed the highest point on its trajectory?
••33 A plane, diving with constant speed at an angle of
53.0° with the vertical, releases a projectile at an altitude of 730 m.
The projectile hits the ground 5.00 s after release. (a) What is the
speed of the plane? (b) How far does the projectile travel horizon-
tally during its flight? What are the (c) horizontal and (d) vertical
components of its velocity just before striking the ground?
••34 A trebuchet was a hurling machine built to attack the
walls of a castle under siege. A large stone could be hurled against a
wall to break apart the wall. The machine was not placed near the
SSM
0
0
θ
d
0
Figure 4-35 Problem 32.
of 9.1 m, its velocity is , with horizontal and
upward. (a) To what maximum height does the ball rise? (b) What
total horizontal distance does the ball travel? What are the
(c) magnitude and (d) angle (below the horizontal) of the ball’s ve-
locity just before it hits the ground?
••44 A baseball leaves a pitcher’s hand horizontally at a speed of
161 km/h.The distance to the batter is 18.3 m. (a) How long does the
ball take to travel the first half of that distance? (b) The second half?
(c) How far does the ball fall freely during the first half? (d) During
the second half? (e) Why aren’t the quantities in (c) and (d) equal?
••45 In Fig. 4-40, a ball is launched with a velocity of magnitude
10.0 m/s, at an angle of 50.0° to the horizontal.The launch point is at
the base of a ramp of horizon-
tal length d16.00 m and
height d23.60 m. A plateau
is located at the top of the
ramp. (a) Does the ball land on
the ramp or the plateau? When
it lands, what are the (b) mag-
nitude and (c) angle of its dis-
placement from the launch point?
••46 In basketball, hang is an illusion in which a player
seems to weaken the gravitational acceleration while in midair.The
illusion depends much on a skilled player’s ability to rapidly shift
j
ˆ
i
ˆ
v
:(7.6i
ˆ6.1j
ˆ) m/s
86 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
wall because then arrows could reach it from the castle wall. Instead,
it was positioned so that the stone hit the wall during the second half
of its flight. Suppose a stone is launched with a speed of v028.0 m/s
and at an angle of u040.0°.What is the speed of the stone if it hits
the wall (a) just as it reaches the top of its parabolic path and (b)
when it has descended to half that height? (c) As a percentage, how
much faster is it moving in part (b) than in part (a)?
••35 A rifle that shoots bullets at 460 m/s is to be aimed at
a target 45.7 m away. If the center of the target is level with the ri-
fle, how high above the target must the rifle barrel be pointed so
that the bullet hits dead center?
••36 During a tennis match, a player serves the ball at
23.6 m/s, with the center of the ball leaving the racquet horizontally
2.37 m above the court surface. The net is 12 m away and 0.90 m
high. When the ball reaches the net, (a) does the ball clear it and
(b) what is the distance between the center of the ball and the top
of the net? Suppose that, instead, the ball is served as before but
now it leaves the racquet at 5.00° below the horizontal. When the
ball reaches the net, (c) does the ball clear it and (d) what now is
the distance between the center of the ball and the top of the net?
••37 A lowly high diver pushes off horizontally
with a speed of 2.00 m/s from the platform edge 10.0 m above the
surface of the water. (a) At what horizontal distance from the
edge is the diver 0.800 s after pushing off? (b) At what vertical
distance above the surface of the water is the diver just then?
(c) At what horizontal distance from the edge does the diver
strike the water?
••38 A golf ball is struck at
ground level. The speed of
the golf ball as a function of
the time is shown in Fig. 4-36,
where t0 at the instant the
ball is struck. The scaling on
the vertical axis is set by
and .
(a) How far does the golf
ball travel horizontally be-
fore returning to ground
level? (b) What is the maximum height above ground level at-
tained by the ball?
••39 In Fig. 4-37, a ball is thrown leftward from the left edge of the
roof, at height habove the ground. The ball hits the ground 1.50 s
later, at distance d25.0 m from the building and at angle u60.0°
with the horizontal. (a) Find h.
(Hint: One way is to reverse the
motion, as if on video.) What
are the (b) magnitude and (c)
angle relative to the horizontal
of the velocity at which the ball
is thrown? (d) Is the angle
above or below the horizontal?
••40 Suppose that a shot putter can put a shot at the world-
class speed 15.00 m/s and at a height of 2.160 m. What hori-
zontal distance would the shot travel if the launch angle is
(a) 45.00° and (b) 42.00°? The answers indicate that the angle of
45°, which maximizes the range of projectile motion, does not max-
imize the horizontal distance when the launch and landing are at
different heights.
0
v0
vb31 m/sva19 m/s
WWWSSM
SSM
18 m
23 m
3.0 m
R
3.0 m
0
v0
θ
Net
Figure 4-39 Problem 42.
v (m/s)
vb
va012
t (s) 345
Figure 4-36 Problem 38.
d
h
θ
Figure 4-37 Problem 39.
••41 Upon spotting an in-
sect on a twig overhanging water, an
archer fish squirts water drops at the
insect to knock it into the water
(Fig. 4-38).Although the fish sees the
insect along a straight-line path at an-
gle fand distance d, a drop must be
launched at a different angle u0if its
parabolic path is to intersect the
insect. If f36.0° and d0.900 m,
what launch angle u0is required for the drop to be at the top of the
parabolic path when it reaches the insect?
••42 In 1939 or 1940, Emanuel Zacchini took his human-
cannonball act to an extreme:After being shot from a cannon, he
soared over three Ferris wheels and into a net (Fig. 4-39). Assume
that he is launched with a speed of 26.5 m/s and at an angle of 53.0°.
(a) Treating him as a particle, calculate his clearance over the first
wheel. (b) If he reached maximum height over the middle wheel, by
how much did he clear it? (c) How far from the cannon should the
net’s center have been positioned (neglect air drag)?
Insect
on twig
d
φ
Archer fish
Figure 4-38 Problem 41.
••43 A ball is shot from the ground into the air. At a height
ILW
Ball
d1
d2
v0
Figure 4-40 Problem 45.
87
PROBLEMS
the ball between hands during the flight, but it might also be sup-
ported by the longer horizontal distance the player travels in the
upper part of the jump than in the lower part. If a player jumps
with an initial speed of v07.00 m/s at an angle of u035.0°,
what percent of the jump’s range does the player spend in the up-
per half of the jump (between maximum height and half maxi-
mum height)?
••47 A batter hits a pitched ball when the center of
the ball is 1.22 m above the ground. The ball leaves the bat at an
angle of 45° with the ground.With that launch,the ball should have
a horizontal range (returning to the launch level) of 107 m. (a)
Does the ball clear a 7.32-m-high fence that is 97.5 m horizontally
from the launch point? (b) At the fence, what is the distance be-
tween the fence top and the ball center?
••48 In Fig. 4-41, a ball is
thrown up onto a roof, landing
4.00 s later at height h20.0 m
above the release level. The
ball’s path just before landing is
angled at u60.0° with the
roof. (a) Find the horizontal dis-
tance dit travels. (See the hint
to Problem 39.) What are the
(b) magnitude and (c) angle
(relative to the horizontal) of
the ball’s initial velocity?
•••49 A football kicker can give the ball an initial speed of
25 m/s. What are the (a) least and (b) greatest elevation angles at
which he can kick the ball to score a field goal from a point 50 m in
front of goalposts whose horizontal bar is 3.44 m above the ground?
•••50 Two seconds after being projected from ground level, a
projectile is displaced 40 m horizontally and 53 m vertically
above its launch point. What are the (a) horizontal and (b)
vertical components of the initial velocity of the projectile? (c)
At the instant the projectile achieves its maximum height above
ground level, how far is it displaced horizontally from the launch
point?
•••51 A skilled skier knows to jump upward before reaching a
downward slope. Consider a jump in which the launch speed is
v010 m/s, the launch angle is u011.3°, the initial course is
approximately flat, and the steeper track has a slope of 9.0°.
Figure 4-42ashows a prejump that allows the skier to land on the top
portion of the steeper track. Figure 4-42bshows a jump at the edge
of the steeper track. In Fig. 4-42a, the skier lands at approximately
the launch level. (a) In the landing, what is the angle fbetween the
skier’s path and the slope? In Fig. 4-42b, (b) how far below the
launch level does the skier land and (c) what is f? (The greater fall
and greater fcan result in loss of control in the landing.)
SSM
WWWSSM
h
d
θ
Figure 4-41 Problem 48.
(a) (b)
Figure 4-42 Problem 51.
•••52 A ball is to be shot from level ground toward a wall at dis-
tance x(Fig. 4-43a). Figure 4-43bshows the ycomponent vyof the
ball’s velocity just as it would reach the wall, as a function of that
vy (m/s)
0
vys
–vys
xs
x (m)
(b)
(a)
y
x
Figure 4-43 Problem 52.
•••53 In Fig. 4-44, a baseball is hit at a height h1.00 m and
then caught at the same height. It travels alongside a wall, moving
up past the top of the wall 1.00 s after it is hit and then down past
the top of the wall 4.00 s later, at distance D50.0 m farther along
the wall. (a) What horizontal distance is traveled by the ball from
hit to catch? What are the (b) magnitude and (c) angle (relative to
the horizontal) of the ball’s velocity just after being hit? (d) How
high is the wall?
D
hh
Figure 4-44 Problem 53.
•••54 A ball is to be shot from
level ground with a certain speed.
Figure 4-45 shows the range Rit will
have versus the launch angle u0.The
value of u0determines the flight
time; let tmax represent the maximum
flight time. What is the least speed
the ball will have during its flight if
u0is chosen such that the flight time
is 0.500tmax?
•••55 A ball rolls horizontally off the top of a stairway with
a speed of 1.52 m/s. The steps are 20.3 cm high and 20.3 cm wide.
Which step does the ball hit first?
Module 4-5 Uniform Circular Motion
•56 An Earth satellite moves in a circular orbit 640 km
(uniform circular motion) above Earth’s surface with a period of
98.0 min. What are (a) the speed and (b) the magnitude of the
centripetal acceleration of the satellite?
•57 A carnival merry-go-round rotates about a vertical axis at a
constant rate. A man standing on the edge has a constant speed of
3.66 m/s and a centripetal acceleration of magnitude 1.83 m/s2.
Position vector locates him relative to the rotation axis. (a) What
is the magnitude of ? What is the direction of when is di-
rected (b) due east and (c) due south?
•58 A rotating fan completes 1200 revolutions every minute.
Consider the tip of a blade, at a radius of 0.15 m. (a) Through what
distance does the tip move in one revolution? What are (b) the
a
:
r
:
r
:
r
:
a
:
SSM
R (m)
200
100
0
0
θ
Figure 4-45 Problem 54.
distance x. The scaling is set by m/s and What
is the launch angle?
xs20 m.vys 5.0
88 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
tip’s speed and (c) the magnitude of its acceleration? (d) What is
the period of the motion?
•59 A woman rides a carnival Ferris wheel at radius 15 m,
completing five turns about its horizontal axis every minute. What
are (a) the period of the motion, the (b) magnitude and (c) direction
of her centripetal acceleration at the highest point, and the (d) mag-
nitude and (e) direction of her centripetal acceleration at the lowest
point?
•60 A centripetal-acceleration addict rides in uniform circular
motion with radius r3.00 m. At one instant his acceleration is
. At that instant, what are the val-
ues of (a) and (b) ?
•61 When a large star becomes a supernova, its core may be
compressed so tightly that it becomes a neutron star, with a radius of
about 20 km (about the size of the San Francisco area). If a neutron
star rotates once every second, (a) what is the speed of a particle on
the star’s equator and (b) what is the magnitude of the particle’s cen-
tripetal acceleration? (c) If the neutron star rotates faster, do the an-
swers to (a) and (b) increase, decrease,or remain the same?
•62 What is the magnitude of the acceleration of a sprinter run-
ning at 10 m/s when rounding a turn of radius 25 m?
••63 At t12.00 s, the acceleration of a particle in counter-
clockwise circular motion is (6.00 m/s2) (4.00 m/s2) . It moves at
constant speed. At time t25.00 s, the particle’s acceleration is
(4.00 m/s2)(6.00 m/s2) . What is the radius of the path taken
by the particle if t2t1is less than one period?
••64 A particle moves horizontally in uniform circular motion,
over a horizontal xy plane. At one instant, it moves through the
point at coordinates (4.00 m, 4.00 m) with a velocity of 5.00 m/s
and an acceleration of 12.5 m/s2. What are the (a) xand (b) y
coordinates of the center of the circular path?
••65 A purse at radius 2.00 m and a wallet at radius 3.00 m travel
in uniform circular motion on the floor of a merry-go-round as the
ride turns. They are on the same radial line. At one instant, the ac-
celeration of the purse is (2.00 m/s2)(4.00 m/s2) .At that instant
and in unit-vector notation, what is the acceleration of the wallet?
••66 A particle moves along a circular path over a horizontal xy
coordinate system, at constant speed.At time t14.00 s, it is at point
(5.00 m, 6.00 m) with velocity (3.00 m/s) and acceleration in the
positive xdirection. At time t210.0 s, it has velocity ( 3.00 m/s)
and acceleration in the positive ydirection. What are the (a) xand
(b) ycoordinates of the center of the circular path if t2t1is less
than one period?
•••67 A boy whirls a stone in a horizontal circle of
radius 1.5 m and at height 2.0 m above level ground. The string
WWWSSM
i
ˆ
j
ˆ
j
ˆ
i
ˆ
j
ˆi
ˆ
j
ˆ
i
ˆ
j
ˆ
i
ˆ
r
:a
:
v
:a
:
a
:(6.00 m/s2)i
ˆ(4.00 m/s2)j
ˆ
ILW
Module 4-6 Relative Motion in One Dimension
•69 A cameraman on a pickup truck is traveling westward at
20 km/h while he records a cheetah that is moving westward
30 km/h faster than the truck. Suddenly, the cheetah stops, turns,
and then runs at 45 km/h eastward, as measured by a suddenly
nervous crew member who stands alongside the cheetah’s path.The
change in the animal’s velocity takes 2.0 s. What are the (a) magni-
tude and (b) direction of the animal’s acceleration according to the
cameraman and the (c) magnitude and (d) direction according to
the nervous crew member?
•70 A boat is traveling upstream in the positive direction of an x
axis at 14 km/h with respect to the water of a river. The water is
flowing at 9.0 km/h with respect to the ground. What are the (a)
magnitude and (b) direction of the boat’s velocity with respect to
the ground? A child on the boat walks from front to rear at
6.0 km/h with respect to the boat.What are the (c) magnitude and
(d) direction of the child’s velocity with respect to the ground?
••71 A suspicious-looking man runs as fast as he can along a
moving sidewalk from one end to the other, taking 2.50 s.Then se-
curity agents appear, and the man runs as fast as he can back along
the sidewalk to his starting point, taking 10.0 s. What is the ratio of
the man’s running speed to the sidewalk’s speed?
Module 4-7 Relative Motion in Two Dimensions
•72 A rugby player runs with the ball directly toward his
opponent’s goal, along the positive direction of an xaxis. He can
legally pass the ball to a teammate as long as the ball’s velocity rela-
tive to the field does not have a positive xcomponent. Suppose the
player runs at speed 4.0 m/s relative to the field while he passes the
ball with velocity relative to himself. If has magnitude
6.0 m/s, what is the smallest angle it can have for the pass to be legal?
••73 Two highways intersect as shown in Fig. 4-46.At the instant
shown, a police car Pis distance dP800 m from the intersection
and moving at speed vP80 km/h. Motorist Mis distance dM
600 m from the intersection and moving at speed vM60 km/h.
v
:
BP
v
:
BP
x
y
M
dM
vP
vM
dP
P
Figure 4-46 Problem 73.
(a) In unit-vector notation, what is the velocity of the motorist
with respect to the police car? (b) For the instant shown in Fig. 4-46,
what is the angle between the velocity found in (a) and the line of
sight between the two cars? (c) If the cars maintain their veloci-
ties, do the answers to (a) and (b) change as the cars move nearer
the intersection?
breaks, and the stone flies off horizontally and strikes the ground
after traveling a horizontal distance of 10 m.What is the magnitude
of the centripetal acceleration of the stone during the circular
motion?
•••68 A cat rides a merry-go-round turning with uniform
circular motion. At time t12.00 s, the cat’s velocity is
, measured on a horizontal xy coordinate
system. At t25.00 s, the cat’s velocity is
. What are (a) the magnitude of the cat’s centripetal
acceleration and (b) the cat’s average acceleration during the time
interval t2t1, which is less than one period?
(4.00 m/s)j
ˆv
:
2(3.00 m/s)i
ˆ
(3.00 m/s)i
ˆ(4.00 m/s)j
ˆv
:
1
89
PROBLEMS
••74 After flying for 15 min in a wind blowing 42 km/h at an
angle of 20° south of east, an airplane pilot is over a town that is
55 km due north of the starting point.What is the speed of the air-
plane relative to the air?
••75 A train travels due south at 30 m/s (relative to the
ground) in a rain that is blown toward the south by the wind. The
path of each raindrop makes an angle of 70° with the vertical, as
measured by an observer stationary on the ground.An observer on
the train, however, sees the drops fall perfectly vertically.
Determine the speed of the raindrops relative to the ground.
••76 A light plane attains an airspeed of 500 km/h. The pilot sets
out for a destination 800 km due north but discovers that the plane
must be headed 20.0° east of due north to fly there directly. The
plane arrives in 2.00 h.What were the (a) magnitude and (b) direc-
tion of the wind velocity?
••77 Snow is falling vertically at a constant speed of 8.0 m/s.
At what angle from the vertical do the snowflakes appear to be
falling as viewed by the driver of a car traveling on a straight, level
road with a speed of 50 km/h?
••78 In the overhead view of
Fig. 4-47, Jeeps Pand Brace
along straight lines, across flat
terrain, and past stationary bor-
der guard A. Relative to the
guard, Btravels at a constant
speed of 20.0 m/s, at the angle
u230.0°. Relative to the guard,
Phas accelerated from rest at a
constant rate of 0.400 m/s2at the
angle u160.0°.At a certain time
during the acceleration, Phas a speed of 40.0 m/s.At that time, what
are the (a) magnitude and (b) direction of the velocity of Prelative to
Band the (c) magnitude and (d) direction of the acceleration of P
relative to B?
••79 Two ships, Aand B, leave port at the same time.
Ship Atravels northwest at 24 knots, and ship Btravels at 28 knots
in a direction 40° west of south. (1 knot 1 nautical mile per hour;
see Appendix D.) What are the (a) magnitude and (b) direction of
the velocity of ship Arelative to B? (c) After what time will the
ships be 160 nautical miles apart? (d) What will be the bearing of B
(the direction of B’s position) relative to Aat that time?
••80 A 200-m-wide river flows due east at a uniform speed of
2.0 m/s. A boat with a speed of 8.0 m/s relative to the water leaves
the south bank pointed in a direction 30° west of north. What are
the (a) magnitude and (b) direction of the boat’s velocity relative
to the ground? (c) How long does the boat take to cross the river?
•••81 Ship Ais located 4.0 km north and 2.5 km east of ship
B. Ship Ahas a velocity of 22 km/h toward the south, and ship B
has a velocity of 40 km/h in a direction 37° north of east. (a)
What is the velocity of Arelative to Bin unit-vector notation
with toward the east? (b) Write an expression (in terms of and )
for the position of Arelative to Bas a function of t, where t0
when the ships are in the positions described above. (c) At what
time is the separation between the ships least? (d) What is that
least separation?
•••82 A 200-m-wide river has a uniform flow speed of 1.1 m/s
through a jungle and toward the east. An explorer wishes to
j
ˆ
i
ˆ
i
ˆ
ILWSSM
SSM
SSM
A
1
θ
2
θ
N
E
P
B
Figure 4-47 Problem 78.
leave a small clearing on the south bank and cross the river in a
powerboat that moves at a constant speed of 4.0 m/s with respect
to the water. There is a clearing on the north bank 82 m up-
stream from a point directly opposite the clearing on the south
bank. (a) In what direction must the boat be pointed in order to
travel in a straight line and land in the clearing on the north
bank? (b) How long will the boat take to cross the river and land
in the clearing?
Additional Problems
83 A woman who can row a boat at 6.4 km/h in still water faces a
long,straight river with a width of 6.4 km and a current of 3.2 km/h.
Let i
ˆpoint directly across the river and ˆ
j point directly down-
stream. If she rows in a straight line to a point directly opposite her
starting position, (a) at what angle to i
ˆmust she point the boat and
(b) how long will she take? (c) How long will she take if, instead,
she rows 3.2 km down the river and then back to her starting
point? (d) How long if she rows 3.2 km up the river and then back
to her starting point? (e) At what angle to ii
ˆshould she point the
boat if she wants to cross the river in the shortest possible time? (f)
How long is that shortest time?
84 In Fig. 4-48a, a sled moves in the negative xdirection at con-
stant speed vswhile a ball of ice is shot from the sled with a velocity
relative to the sled. When the ball lands, its hori-
zontal displacement xbg relative to the ground (from its launch
position to its landing position) is measured. Figure 4-48bgives
xbg as a function of vs. Assume the ball lands at approximately
its launch height. What are the values of (a) v0xand (b) v0y? The
ball’s displacement xbs relative to the sled can also be measured.
Assume that the sled’s velocity is not changed when the ball is
shot.What is xbs when vsis (c) 5.0 m/s and (d) 15 m/s?
v
:
0v0xi
ˆv0yj
ˆ
Figure 4-48 Problem 84.
Ball
Sled
y
x
vs
(a)
(b)
10
0
–40
40
20
Δxbg (m)
vs (m/s)
85 You are kidnapped by political-science majors (who are
upset because you told them political science is not a real
science). Although blindfolded, you can tell the speed of their
car (by the whine of the engine), the time of travel (by mentally
counting off seconds), and the direction of travel (by turns
along the rectangular street system). From these clues, you
know that you are taken along the following course: 50 km/h for
2.0 min, turn 90° to the right, 20 km/h for 4.0 min, turn 90° to the
right, 20 km/h for 60 s, turn 90° to the left, 50 km/h for 60 s, turn
90° to the right, 20 km/h for 2.0 min, turn 90° to the left, 50 km/h
for 30 s. At that point, (a) how far are you from your starting
point, and (b) in what direction relative to your initial direction
of travel are you?
87 A baseball is hit at ground level. The ball reaches its
maximum height above ground level 3.0 s after being hit. Then
2.5 s after reaching its maximum height, the ball barely clears a
fence that is 97.5 m from where it was hit. Assume the ground is
level. (a) What maximum height above ground level is reached by
the ball? (b) How high is the fence? (c) How far beyond the fence
does the ball strike the ground?
88 Long flights at midlatitudes in the Northern Hemisphere en-
counter the jet stream, an eastward airflow that can affect a plane’s
speed relative to Earth’s surface. If a pilot maintains a certain speed
relative to the air (the plane’s airspeed),the speed relative to the sur-
face (the plane’s ground speed) is more when the flight is in the di-
rection of the jet stream and less when the flight is opposite the jet
stream. Suppose a round-trip flight is scheduled between two cities
separated by 4000 km, with the outgoing flight in the direction of the
jet stream and the return flight opposite it.The airline computer ad-
vises an airspeed of 1000 km/h, for which the difference in flight
times for the outgoing and return flights is 70.0 min.What jet-stream
speed is the computer using?
89 A particle starts from the origin at t0 with a velocity
of 8.0 m/s and moves in the xy plane with constant acceleration
(4.0 2.0 ) m/s2. When the particle’s xcoordinate is 29 m, what
are its (a) ycoordinate and (b) speed?
90 At what initial speed
must the basketball player in
Fig. 4-50 throw the ball, at an-
gle u055° above the hori-
zontal, to make the foul shot?
The horizontal distances are
d11.0 ft and d214 ft, and
the heights are h17.0 ft
and h210 ft.
91 During volcanic erup-
tions, chunks of solid rock
can be blasted out of the vol-
cano; these projectiles are
called volcanic bombs. Figure 4-51 shows a cross section of Mt.
Fuji, in Japan. (a) At what initial speed would a bomb have to be
ejected, at angle u035° to the horizontal, from the vent at Ain
order to fall at the foot of the volcano at B, at vertical distance
h3.30 km and horizontal distance d9.40 km? Ignore, for the
j
ˆ
i
ˆ
j
ˆ
SSM
SSM
θ
h
d
A
B
0
90 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
86 A radar station detects an airplane approaching directly from
the east. At first observation, the airplane is at distance d1360 m
from the station and at angle u140° above the horizon (Fig. 4-49).
The airplane is tracked through an angular change u123° in the
vertical east–west plane; its distance is then d2790 m. Find the
(a) magnitude and (b) direction of the airplane’s displacement dur-
ing this period.
h2
h1
0
d1
d2
θ
Figure 4-50 Problem 90.
moment, the effects of air on the bomb’s travel. (b) What would
be the time of flight? (c) Would the effect of the air increase or
decrease your answer in (a)?
Figure 4-51 Problem 91.
y
x (km)
You
20100
Figure 4-52 Problem 94.
92 An astronaut is rotated in a horizontal centrifuge at a radius
of 5.0 m. (a) What is the astronaut’s speed if the centripetal accel-
eration has a magnitude of 7.0g? (b) How many revolutions per
minute are required to produce this acceleration? (c) What is the
period of the motion?
93 Oasis Ais 90 km due west of oasis B. A desert camel
leaves Aand takes 50 h to walk 75 km at 37° north of due east.
Next it takes 35 h to walk 65 km due south. Then it rests for 5.0 h.
What are the (a) magnitude and (b) direction of the camel’s dis-
placement relative to Aat the resting point? From the time the
camel leaves Auntil the end of the rest period, what are the (c)
magnitude and (d) direction of its average velocity and (e) its aver-
age speed? The camel’s last drink was at A; it must be at Bno more
than 120 h later for its next drink. If it is to reach Bjust in time, what
must be the (f) magnitude and (g) direction of its average velocity
after the rest period?
94 Curtain of death. A large metallic asteroid strikes Earth
and quickly digs a crater into the rocky material below ground level
by launching rocks upward and outward. The following table gives
five pairs of launch speeds and angles (from the horizontal) for such
rocks, based on a model of crater formation. (Other rocks, with inter-
mediate speeds and angles, are also launched.) Suppose that you are
at x20 km when the asteroid strikes the ground at time t0 and
position x0 (Fig. 4-52). (a) At t20 s, what are the xand y
coordinates of the rocks headed in your direction from launches A
through E? (b) Plot these coordinates and then sketch a curve
through the points to include rocks with intermediate launch speeds
and angles.The curve should indicate what you would see as you look
up into the approaching rocks.
Launch Speed (m/s) Angle (degrees)
A520 14.0
B630 16.0
C750 18.0
D870 20.0
E1000 22.0
SSM
E
W
d2
Airplane
Radar dish
1
θ
Δ
θ
d1
Figure 4-49 Problem 86.
91
PROBLEMS
95 Figure 4-53 shows the straight path of a particle
across an xy coordinate system as the particle is ac-
celerated from rest during time interval t1. The ac-
celeration is constant. The xy coordinates for point
Aare (4.00 m, 6.00 m); those for point Bare (12.0
m, 18.0 m). (a) What is the ratio ay/axof the acceler-
ation components? (b) What are the coordinates of
the particle if the motion is continued for another
interval equal to t1?
96 For women’s volleyball the top of the net is 2.24 m above the
floor and the court measures 9.0 m by 9.0 m on each side of the
net. Using a jump serve, a player strikes the ball at a point that is
3.0 m above the floor and a horizontal distance of 8.0 m from the
net. If the initial velocity of the ball is horizontal, (a) what mini-
mum magnitude must it have if the ball is to clear the net and (b)
what maximum magnitude can it have if the ball is to strike the
floor inside the back line on the other side of the net?
97 A rifle is aimed horizontally at a target 30 m away. The
bullet hits the target 1.9 cm below the aiming point.What are (a) the
bullet’s time of flight and (b) its speed as it emerges from the rifle?
98 A particle is in uniform circular motion about the origin of an
xy coordinate system, moving clockwise with a period of 7.00 s. At
one instant, its position vector (measured from the origin) is
. At that instant, what is its velocity in
unit-vector notation?
99 In Fig. 4-54, a lump of wet
putty moves in uniform circular mo-
tion as it rides at a radius of 20.0 cm
on the rim of a wheel rotating coun-
terclockwise with a period of 5.00
ms. The lump then happens to fly off
the rim at the 5 o’clock position (as
if on a clock face). It leaves the rim
at a height of h1.20 m from the floor and at a distance d2.50
m from a wall.At what height on the wall does the lump hit?
100 An iceboat sails across the surface of a frozen lake with con-
stant acceleration produced by the wind. At a certain instant the
boat’s velocity is (6.30 8.42 ) m/s. Three seconds later, because
of a wind shift, the boat is instantaneously at rest. What is its aver-
age acceleration for this 3.00 s interval?
101 In Fig. 4-55, a ball is shot di-
rectly upward from the ground with
an initial speed of v07.00 m/s.
Simultaneously, a construction eleva-
tor cab begins to move upward from
the ground with a constant speed of
vc3.00 m/s. What maximum height
does the ball reach relative to (a) the
ground and (b) the cab floor? At what rate does the speed of the ball
change relative to (c) the ground and (d) the cab floor?
102 A magnetic field forces an electron to move in a circle with
radial acceleration 3.0 1014 m/s2. (a) What is the speed of the elec-
tron if the radius of its circular path is 15 cm? (b) What is the period
of the motion?
103 In 3.50 h, a balloon drifts 21.5 km north, 9.70 km east, and
2.88 km upward from its release point on the ground. Find (a) the
magnitude of its average velocity and (b) the angle its average ve-
locity makes with the horizontal.
j
ˆ
i
ˆ
r
:(2.00 m)i
ˆ(3.00 m)j
ˆ
SSM
h
d
Putty
Wheel
Figure 4-54 Problem 99.
vc
v0
Ball
Figure 4-55 Problem 101.
104 A ball is thrown horizontally from a height of 20 m and hits
the ground with a speed that is three times its initial speed.What is
the initial speed?
105 A projectile is launched with an initial speed of 30 m/s at an
angle of 60° above the horizontal.What are the (a) magnitude and
(b) angle of its velocity 2.0 s after launch, and (c) is the angle above
or below the horizontal? What are the (d) magnitude and (e) angle
of its velocity 5.0 s after launch, and (f) is the angle above or below
the horizontal?
106 The position vector for a proton is initially
and then later is , all
in meters. (a) What is the proton’s displacement vector, and (b) to
what plane is that vector parallel?
107 A particle Ptravels with con-
stant speed on a circle of radius r
3.00 m (Fig. 4-56) and completes one
revolution in 20.0 s. The particle
passes through Oat time t0. State
the following vectors in magnitude-
angle notation (angle relative to the
positive direction of x). With respect
to O, find the particle’s position vec-
tor at the times tof (a) 5.00 s, (b)
7.50 s, and (c) 10.0 s. (d) For the
5.00 s interval from the end of
the fifth second to the end of the
tenth second, find the particle’s displacement. For that interval,
find (e) its average velocity and its velocity at the (f) beginning and
(g) end. Next, find the acceleration at the (h) beginning and (i) end
of that interval.
108 The fast French train known as the TGV (Train à Grande
Vitesse) has a scheduled average speed of 216 km/h. (a) If the train
goes around a curve at that speed and the magnitude of the accel-
eration experienced by the passengers is to be limited to 0.050g,
what is the smallest radius of curvature for the track that can be
tolerated? (b) At what speed must the train go around a curve with
a 1.00 km radius to be at the acceleration limit?
109 (a) If an electron is projected horizontally with a speed of
3.0 106m/s, how far will it fall in traversing 1.0 m of horizontal
distance? (b) Does the answer increase or decrease if the initial
speed is increased?
110 A person walks up a stalled 15-m-long escalator in 90 s.
When standing on the same escalator, now moving, the person is
carried up in 60 s. How much time would it take that person to
walk up the moving escalator? Does the answer depend on the
length of the escalator?
111 (a) What is the magnitude of the centripetal acceleration of
an object on Earth’s equator due to the rotation of Earth? (b)
What would Earth’s rotation period have to be for objects on the
equator to have a centripetal acceleration of magnitude 9.8 m/s2?
112 The range of a projectile depends not only on v0and
but also on the value gof the free-fall acceleration, which varies
from place to place. In 1936, Jesse Owens established a world’s
running broad jump record of 8.09 m at the Olympic Games at
Berlin (where g9.8128 m/s2). Assuming the same values of v0
and , by how much would his record have differed if he had com-
peted instead in 1956 at Melbourne (where g9.7999 m/s2)?
0
0
r
:2.0i
ˆ6.0j
ˆ2.0k
ˆ
5.0i
ˆ6.0j
ˆ2.0k
ˆr
:
x
rP
y
O
Figure 4-56 Problem 107.
y
x
A
B
Figure 4-53
Problem 95.
92 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS
113 Figure 4-57 shows the path
taken by a drunk skunk over level
ground, from initial point ito final
point f. The angles are 30.0°,
50.0°, and 80.0°, and the
distances are d15.00 m, d28.00
m, and d312.0 m.What are the (a)
magnitude and (b) angle of the
skunk’s displacement from ito f?
114 The position vector of a
particle moving in the xy plane is
, with
in meters and tin seconds. (a)
Calculate the xand ycomponents
of the particle’s position at t0, 1.0, 2.0, 3.0, and 4.0 s and
sketch the particle’s path in the xy plane for the interval 0 t
4.0 s. (b) Calculate the components of the particle’s velocity at
t1.0, 2.0, and 3.0 s. Show that the velocity is tangent to the
path of the particle and in the direction the particle is moving at
each time by drawing the velocity vectors on the plot of the parti-
cle’s path in part (a). (c) Calculate the components of the parti-
cle’s acceleration at t1.0, 2.0, and 3.0 s.
115 An electron having an initial horizontal velocity of magnitude
1.00 109cm/s travels into the region between two horizontal metal
plates that are electrically charged. In that region, the electron trav-
els a horizontal distance of 2.00 cm and has a constant downward ac-
celeration of magnitude 1.00 1017 cm/s2due to the charged plates.
Find (a) the time the electron takes to travel the 2.00 cm, (b) the ver-
tical distance it travels during that time, and the magnitudes of its (c)
horizontal and (d) vertical velocity components as it emerges from
the region.
116 An elevator without a ceiling is ascending with a constant
speed of 10 m/s. A boy on the elevator shoots a ball directly up-
ward, from a height of 2.0 m above the elevator floor, just as the el-
evator floor is 28 m above the ground.The initial speed of the ball
with respect to the elevator is 20 m/s. (a) What maximum height
above the ground does the ball reach? (b) How long does the ball
take to return to the elevator floor?
117 A football player punts the football so that it will have a
“hang time” (time of flight) of 4.5 s and land 46 m away. If the ball
leaves the player’s foot 150 cm above the ground, what must be the
(a) magnitude and (b) angle (relative to the horizontal) of the
ball’s initial velocity?
118 An airport terminal has a moving sidewalk to speed passen-
gers through a long corridor. Larry does not use the moving side-
walk; he takes 150 s to walk through the corridor. Curly, who sim-
ply stands on the moving sidewalk, covers the same distance in 70 s.
Moe boards the sidewalk and walks along it. How long does Moe
take to move through the corridor? Assume that Larry and Moe
walk at the same speed.
119 A wooden boxcar is moving along a straight railroad track
at speed v1. A sniper fires a bullet (initial speed v2) at it from a
high-powered rifle. The bullet passes through both lengthwise
walls of the car, its entrance and exit holes being exactly opposite
each other as viewed from within the car. From what direction, rel-
ative to the track, is the bullet fired? Assume that the bullet is not
deflected upon entering the car, but that its speed decreases by
20%.Take v185 km/h and v2650 m/s. (Why don’t you need to
know the width of the boxcar?)
r
:
r
:2ti
ˆ2 sin[(
/4 rad/s)t]j
ˆ
r
:
3
2
1
120 A sprinter running on a circular track has a velocity of con-
stant magnitude 9.20 m/s and a centripetal acceleration of magni-
tude 3.80 m/s2. What are (a) the track radius and (b) the period of
the circular motion?
121 Suppose that a space probe can withstand the stresses of a
20gacceleration. (a) What is the minimum turning radius of such a
craft moving at a speed of one-tenth the speed of light? (b) How
long would it take to complete a 90° turn at this speed?
122 You are to throw a ball with
a speed of 12.0 m/s at a target that is
height h= 5.00 m above the level at
which you release the ball (Fig. 4-58).
You want the ball’s velocity to be
horizontal at the instant it reaches
the target. (a) At what angle above
the horizontal must you throw the
ball? (b) What is the horizontal dis-
tance from the release point to the
target? (c) What is the speed of the
ball just as it reaches the target?
123 A projectile is fired with an
initial speed v0= 30.0 m/s from level
ground at a target that is on the
ground, at distance R= 20.0 m, as
shown in Fig. 4-59. What are the (a)
least and (b) greatest launch angles
that will allow the projectile to hit the
target?
124 A graphing surprise.At time t= 0, a burrito is launched from
level ground, with an initial speed of 16.0 m/s and launch angle .
Imagine a position vector continuously directed from the
launching point to the burrito during the flight. Graph the magni-
tude rof the position vector for (a) = 40.0° and (b) = 80.0°. For
= 40.0°, (c) when does rreach its maximum value, (d) what is
that value, and how far (e) horizontally and (f) vertically is the bur-
rito from the launch point? For = 80.0°, (g) when does rreach its
maximum value, (h) what is that value, and how far (i) horizontally
and (j) vertically is the burrito from the launch point?
125 A cannon located at sea level fires a ball with initial speed
82 m/s and initial angle 45°.The ball lands in the water after travel-
ing a horizontal distance 686 m. How much greater would the hori-
zontal distance have been had the cannon been 30 m higher?
126 The magnitude of the velocity of a projectile when it is at its
maximum height above ground level is 10.0 m/s. (a) What is the
magnitude of the velocity of the projectile 1.00 s before it achieves
its maximum height? (b) What is the magnitude of the velocity of
the projectile 1.00 s after it achieves its maximum height? If we
take x= 0 and y= 0 to be at the point of maximum height and posi-
tive xto be in the direction of the velocity there, what are the (c) x
coordinate and (d) ycoordinate of the projectile 1.00 s before it
reaches its maximum height and the (e) xcoordinate and (f) yco-
ordinate 1.0 s after it reaches its maximum height?
127 A frightened rabbit moving at 6.00 m/s due east runs onto a
large area of level ice of negligible friction. As the rabbit slides
across the ice, the force of the wind causes it to have a constant ac-
celeration of 1.40 m/s2, due north. Choose a coordinate system with
the origin at the rabbit’s initial position on the ice and the positive
xaxis directed toward the east. In unit-vector notation, what are
the rabbit’s (a) velocity and (b) position when it has slid for 3.00 s?
0
0
0
0
r
:
0
h
Target
θ
Figure 4-58 Problem 122.
v0
v0
R
High trajectory
Low trajectory
Figure 4-59 Problem 123.
y
x
θ
3
θ
1
θ
2
d
3
f
i
d
2
d
1
Figure 4-57 Problem 113.
93
PROBLEMS
130 Some state trooper departments use aircraft to enforce
highway speed limits. Suppose that one of the airplanes has a speed
of 135 mi/h in still air. It is flying straight north so that it is at all
times directly above a north–south highway. A ground observer
tells the pilot by radio that a 70.0 mi/h wind is blowing but neglects
to give the wind direction. The pilot observes that in spite of the
wind the plane can travel 135 mi along the highway in 1.00 h. In
other words, the ground speed is the same as if there were no wind.
(a) From what direction is the wind blowing? (b) What is the head-
ing of the plane; that is, in what direction does it point?
131 A golfer tees off from the top of a rise, giving the golf ball an
initial velocity of 43.0 m/s at an angle of 30.0° above the horizontal.
The ball strikes the fairway a horizontal distance of 180 m from the
tee. Assume the fairway is level. (a) How high is the rise above the
fairway? (b) What is the speed of the ball as it strikes the fairway?
132 A track meet is held on a planet in a distant solar system. A
shot-putter releases a shot at a point 2.0 m above ground level. A
stroboscopic plot of the position of the shot is shown in Fig. 4-61,
where the readings are 0.50 s apart and the shot is released at
time t= 0. (a) What is the initial velocity of the shot in unit-vector
notation? (b) What is the magnitude of the free-fall acceleration
on the planet? (c) How long after it is released does the shot
reach the ground? (d) If an identical throw of the shot is made on
the surface of Earth, how long after it is released does it reach the
ground?
133 A helicopter is flying in a straight line over a level field at
a constant speed of 6.20 m/s and at a constant altitude of 9.50 m.
A package is ejected horizontally from the helicopter with an
initial velocity of 12.0 m/s relative to the helicopter and in a di-
rection opposite the helicopter’s motion. (a) Find the initial
speed of the package relative to the ground. (b) What is the hori-
zontal distance between the helicopter and the package at the
instant the package strikes the ground? (c) What angle does the
velocity vector of the package make with the ground at the in-
stant before impact, as seen from the ground?
134 A car travels around a flat circle on the ground, at a constant
speed of 12.0 m/s.At a certain instant the car has an acceleration of
3.00 m/s2toward the east. What are its distance and direction from
the center of the circle at that instant if it is traveling (a) clockwise
around the circle and (b) counterclockwise around the circle?
135 You throw a ball from a cliff with an initial velocity of
15.0 m/s at an angle of 20.0° below the horizontal. Find (a) its hori-
zontal displacement and (b) its vertical displacement 2.30 s later.
136 A baseball is hit at Fenway Park in Boston at a point
0.762 m above home plate with an initial velocity of 33.53 m/s di-
rected 55.0° above the horizontal. The ball is observed to clear
the 11.28-m-high wall in left field (known as the “green mon-
ster”) 5.00 s after it is hit, at a point just inside the left-field foul-
line pole. Find (a) the horizontal distance down the left-field foul
line from home plate to the wall; (b) the vertical distance by
which the ball clears the wall; (c) the horizontal and vertical dis-
placements of the ball with respect to home plate 0.500 s before
it clears the wall.
137 A transcontinental flight of 4350 km is scheduled to take
50 min longer westward than eastward. The airspeed of the air-
plane is 966 km/h, and the jet stream it will fly through is pre-
sumed to move due east. What is the assumed speed of the jet
stream?
138 A woman can row a boat at 6.40 km/h in still water. (a) If
she is crossing a river where the current is 3.20 km/h, in what di-
rection must her boat be headed if she wants to reach a point di-
rectly opposite her starting point? (b) If the river is 6.40 km
wide, how long will she take to cross the river? (c) Suppose that
instead of crossing the river she rows 3.20 km down the river and
then back to her starting point. How long will she take? (d) How
long will she take to row 3.20 km up the river and then back to
her starting point? (e) In what direction should she head the
boat if she wants to cross in the shortest possible time, and what
is that time?
y (ft)
10
5
01020
x (ft)
30 40
t = 0
Figure 4-60 Problem 129.
y (m)
10
5
01052015 25
x (m)
30
t = 0
Figure 4-61 Problem 132.
128 The pilot of an aircraft flies due east relative to the ground
in a wind blowing 20.0 km/h toward the south. If the speed of the
aircraft in the absence of wind is 70.0 km/h, what is the speed of the
aircraft relative to the ground?
129 The pitcher in a slow-pitch softball game releases the ball at a
point 3.0 ft above ground level.A stroboscopic plot of the position of
the ball is shown in Fig. 4-60, where the readings are 0.25 s apart and
the ball is released at t= 0. (a) What is the initial speed of the ball?
(b) What is the speed of the ball at the instant it reaches its maxi-
mum height above ground level? (c) What is that maximum height?
